if such three condition find the $\sin{\angle OPA}$

$$\textbf{Edition of 11.01.2019}$$

$\color{brown}{\textbf{The issue data}}$

Let $$\angle OPA=\varphi,\quad \angle DAB = \alpha,\quad \angle CDA=\delta,\quad AD=d=13x,\quad PB=u,\quad PC=v.\tag1$$ Denote \begin{vmatrix} AB & BC & CD & AC & BD & PA & PD & AO\\ a & b & c & f & g & - & - & R\\ \dfrac{8x+u}3 & \dfrac{u+v}2 & \dfrac23(5x+v) & - & - & 8x & 5x & R\\ \end{vmatrix}

$$ $$ $\color{brown}{\textbf{Triangles AOP and POD}}$

$$AO = OD = R,\quad AP=8x,\quad PD=5x,\quad \angle OPA=\varphi.$$ Let $$OP=y,\quad \angle OAP = \angle ODP = \psi.$$ Using cosine theorem, one can get $$ \begin{cases} y^2 = (8x)^2 + R^2 - 16Rx\cos\psi\\[4pt] y^2 = (5x)^2 + R^2 - 10Rx\cos\psi\\[4pt] R^2 = (5x)^2+y^2+10xy\cos\varphi \end{cases}\rightarrow \begin{cases} y^2=R^2-40x^2,\\[4pt] \cos\varphi = \dfrac{R^2-y^2-25x^2}{10xy}, \end{cases} $$ $$\cos\varphi = \dfrac{3x}{2\sqrt{R^2-40x^2}}.\tag2$$

$\color{brown}{\textbf{Triangle ABP}}$ $$\angle DAB = \alpha,\quad AB=a=\dfrac{8x+u}3,\quad PA=8x,\quad PB=u.$$ Cosine theorem leads to the equation $$u^2=a^2+(8x)^2-16ax\cos\alpha,$$ $$\cos\alpha = \dfrac{a^2+(8x)^2-u^2}{16ax}=\dfrac{(8x+u)^2+9(8x-u)(8x+u)}{48x(8x+u)} =\dfrac{8x+u+9(8x-u)}{48x},$$ $$\cos\alpha = \dfrac{10x-u}{6x},\quad \sin\alpha = \dfrac1{6x}\sqrt{(6x)^2-(10x-u)^2}\tag3.$$

$\color{brown}{\textbf{Triangles ABD and BCD}}$

$$\angle DAB = \alpha,\quad AB=a=\dfrac{8x+u}3,\quad AD=13x,\quad BC=b=\dfrac{u+v}2,\quad CD=c=\dfrac23(5x+v),$$ In the inscribed quadrilateral, the sums of opposite angles are equal to $\pi,$ so $$\angle BCD = \pi-\alpha,\quad \cos(\angle BCD) = -\cos\alpha.$$ Cosine theorem for $\triangle ABD$ and $\triangle BCD$ gives $$f^2 = a^2+d^2-2ad\cos\alpha = b^2+c^2+2bc\cos\alpha,$$ $$f^2=\dfrac19((8x+u)^2+(39x)^2 - 13(8x+u)(10x-u)) = \dfrac1{36x}\left(9x(u+v)^2+16x(5x+v)^2+4(u+v)(5x+v)(10x-u)\right),$$ $$f^2=\dfrac19(14u^2-10ux+545x^2),\tag4$$ $$u^2 (4 v + 67 x) + 2 u (2 v^2 - 19 v x - 120 x^2) - 5 x (13 v^2 + 72 v x - 356 x^2) = 0.\tag5$$

$\triangle ABD$ is inscribed to the same circle as ABCD. Sine theorem gives $$R = \dfrac f{2\sin\alpha}.$$ Using $(3)-(4),$ $$R = x\sqrt{\dfrac{14u^2-10ux+545x^2}{36x^2-(10x-u)^2}}.\tag6$$

$\color{brown}{\textbf{Triangle PCD}}$ $$\angle PDC = \delta,\quad CD=c=\dfrac{10x+2v}3,\quad PD=5x,\quad PC=v.$$

Cosine theorem leads to the equation $$v^2=c^2+(5x)^2-10cx\cos\delta,$$ $$\cos\delta = \dfrac{c^2+(5x)^2-v^2}{10cx}=\dfrac{4(5x+v)^2+9(5x-v)(5x+v)}{60x(5x+v)},$$ $$\cos\delta = \dfrac{13x-v}{12x},\quad \sin\delta = \dfrac1{12x}\sqrt{(12x)^2-(13x-v)^2}.\tag7$$

$\color{brown}{\textbf{Triangles ACD and ABC}}$

$$\angle ADC = \delta,\quad AB=a=\dfrac{8x+u}3,\quad BC=b=\dfrac{u+v}2,\quad CD=c=\dfrac{10x+2v}3,\quad AD=d=13x.$$ In the inscribed quadrilateral, the sums of opposite angles are equal to $\pi,$ so $$\angle ABC = \pi-\delta,\quad \cos(\angle ABD) = -\cos\delta.$$ Cosine theorem for $\triangle ACD$ and $\triangle ABC$ gives $$g^2 = c^2+d^2-2cd\cos\delta = a^2+b^2+2ab\cos\delta,$$ $$g^2=\dfrac19((10x+2v)^2+(39x)^2 - 13(5x+v)(13x-v)) = \dfrac1{36x}\left(4x(8x+u)^2+9x(u+v)^2+(8x+u)(u+v)(13x-v)\right),$$ $$g^2=\dfrac19(17v^2-64vx+776x^2),\tag8$$ $$u^2 v - 26 u^2 x + u v^2 - 23 u v x - 168 u x^2 + 67 v^2 x - 360 v x^2 + 2848 x^3 =0.\tag9$$

$\triangle ACD$ is inscribed to the same circle as ABCD. Sine theorem gives $$R = \dfrac g{2\sin\delta}.$$ Using $(3)-(4),$ $$R = x\sqrt{\dfrac{68v^2-256vx+3104x^2}{144x^2-(13x-v)^2}}.\tag{10}$$

$\color{brown}{\textbf{Constraints to the radius}}$

Equations $(6),(10)$ form the system $$R^2 = \dfrac{14u^2-10u+545}{-u^2+20u-64}=\dfrac{68v^2-256v+3104}{-v^2+26v-25}.\tag{11}$$

Easy to show that $$\min\limits_{u\in(4,16)} R^2 = \dfrac{169}4\quad\text{at}\quad u=\dfrac{38}5$$ (see also Wolfram Alpha), $$\min\limits_{v\in(1,25)} R^2 = \dfrac{169}4\quad\text{at}\quad v=\dfrac{43}7$$ (see also Wolfram Alpha). I.e. the least value of $R$ corresponds to the trivial case $AD=2R.$

At the same time, using of $(11)$ in the calculations of $u$ and $v$ is not suitable.

$\color{brown}{\textbf{Calculation of u and v}}$

If assume $x=1$, then will be obtained solution for $\dfrac ux,\ \dfrac vx.$ Equations $(4),(9)$ under this condition form the system $$\begin{cases} u^2(4v+67)+u(4v^2-38v-240)-65v^2-360v+1780 = 0\\ u^2(v-26) + u(v^2-23v-168)+ 67v^2-360v+2848=0. \end{cases}\tag{12}$$ Summation of equations $(12)$ with factors $1$ and $(-4)$ gives $$\begin{cases} 19u^2+6uv-37v^2+48u+120v-1068 = 0,\\ u^2(v-26) + u(v^2-23v-168)+ 67v^2-360v+2848=0. \end{cases}\tag{13}$$ Taking in account that $$19(u^2(v-26) + u(v^2-23v-168)+ 67v^2-360v+2848)\\ =(v-26)(19u^2+6uv-37v^2+48u+120v)+(13v^2-329v-1944)u\\ + 37v^3+191v^2-3720v+54112,$$ this leads to the system with the explicit expresssion for $p,$ $$\begin{cases} 19u^2+6uv-37v^2+48u+120v-1068 = 0,\\ u(13v^2-329v-1944)+37v^3+191v^2-2652v+26344=0, \end{cases}\tag{14}$$ so $$\begin{cases} u=\dfrac{37 v^3 + 191 v^2 - 2652 v + 26344}{-13 v^2 + 329 v + 1944}\\[4pt] 222v^6+8426v^5-64431v^4-137056v^3+3387595v^2-40679520v+152739664=0\end{cases}\tag{15}$$ wherein $$222=2\cdot3\cdot37,\quad 152739664 = 2^4\cdot7^2\cdot11\cdot89\cdot199.\tag{16}$$ The integer roots are dividers of $152739664.$ Horner's scheme for them is \begin{vmatrix} v & \mathbf{222} & \mathbf{8426} & \mathbf{-64431} & \mathbf{-137056} & \mathbf{3387595} & \mathbf{-40679520} & \mathbf{152739664}\\ 2 & 222 & 8870 & -46691 & -230438 & 2926719 & -34826082 & 83087500\\ -2 & 222 & 7982 & -80395 & 23734 & 3340127 & -47359774 & 247459212 \\ 4 & 222 & 9314 & -27175 & -245756 & 2404571 & -31061236 & 28494720\\ -4 & 222 & 7538 & -94583 & 241276 & 2422491 & -50369484 & 354217600\\ \color{brown}{\mathbf{7}} & \mathbf{222} & \mathbf{9980} & \mathbf{5429} & \mathbf{-99053} & \mathbf{2694224} & \mathbf{-21819952} & 0 \\ \end{vmatrix} The root $v=7$ is obtained, and coefficients of the reduced polynomial are obtained too.

After the integer roots should be checked rational roots, using dividers of 222 as denominators.

Results of the searching allow to write the system in the form of $$\begin{cases} (v - 7) (v + 44) (3 v + 28) (37 v - 199) (2 v^2 - 6 v + 89) = 0\\[4pt] u=\dfrac{37 v^3 + 191 v^2 - 2652 v + 26344}{-13 v^2 + 329 v + 1944}, \end{cases}$$ with the positive solutions $$\binom{u/x}{v/x} = \left\{\binom{157/19}7, \binom7{199/37}\right\}.\tag{17}$$

Obtained values $(17)$ allow to obtain full solutions, using $(6),(2):$ $$\left(\dfrac Rx\right)^2 = \dfrac{14\left(\dfrac ux\right)^2-10\left(\dfrac ux\right)+545}{36-\left(\left(\dfrac ux\right)-10\right)^2},\tag{18}$$ $$\cos\varphi = \dfrac{3}{2\sqrt{\left(\dfrac Rx\right)^2-40}}.\tag{19}$$ From $(18)$ should $$\dfrac ux \in(4,16).$$

$\color{brown}{\textbf{Results}}$

If $\dfrac ux = 7,$ then $$\left(\dfrac Rx\right)^2=43,\quad \cos\varphi = \dfrac{\sqrt3}2,$$ $$\color{brown}{\boxed{\mathbf{\varphi = \dfrac\pi6}}}.$$

If $\dfrac ux=\frac{157}{19}\approx 8.26316$ then $$\left(\dfrac Rx\right)^2=43,\quad \cos\varphi = \dfrac{\sqrt3}2,$$ $$\color{brown}{\boxed{\mathbf{\varphi = \dfrac\pi6}}}.$$

Therefore, both of the possible solutions leads to the similar angle $\varphi.$

$\color{brown}{\textbf{Coordinates}}$

If $O=(0,0),\quad x=1,\quad u=7,\quad v=\dfrac{199}{37},\quad R=\sqrt{43},\quad b=\dfrac{229}{37},\quad c=\dfrac{256}{37},\quad$ then \begin{vmatrix} \mathbf{a} & \mathbf{\cos\alpha} & \mathbf{\sin\alpha} & \mathbf{A_X} & \mathbf{B_X} & \mathbf{B_Y} & \mathbf{A_Y} \\ \dfrac{8x+u}3 & \dfrac{10x-u}{6x} & \sqrt{1-\cos^2\alpha} & -\dfrac{13}2x & A_X+a\cos\alpha & \sqrt{R^2-B_X^2} & B_Y-a\sin\alpha \\ 5 & \dfrac12 & \dfrac{\sqrt3}2 & -\dfrac{13}2 & -4 & 3\sqrt3 &\dfrac{\sqrt3}2\\ \end{vmatrix} The other parameters are \begin{vmatrix} \mathbf{c} & \mathbf{\cos\delta} & \mathbf{\sin\delta} & \mathbf{P} & \mathbf{D} & \mathbf{C} \\ \dfrac{2(5x+v)}3 & \dfrac{13x-v}{12x} & \sqrt{1-\cos^2\delta} & \left(\dfrac32 x,A_Y\right) & \left(\dfrac{13}2 x, A_Y\right) & D + c(-\cos\delta, \sin\delta) \\ \dfrac{256}{37} & \dfrac{47}{74} & \dfrac{33}{74}\sqrt3 & \left(\dfrac32,\dfrac{\sqrt3}2\right) & \left(\dfrac{13}2,\dfrac{\sqrt3}2\right) & \left(\dfrac{5765}{2738},\dfrac{9817}{2738}\sqrt3\right) \\ \end{vmatrix}

One can check that $$\|A\| = \|B\| = \|C\| = \|D\| = \sqrt{43},\quad \tan\dfrac{P_Y}{P_X}=\dfrac{\sqrt3}3.$$

If $O=(0,0),\quad x=1,\quad u=\dfrac{157}{19},\quad v=7,\quad R=\sqrt{43},\quad b=\dfrac{145}{19},\quad c = 8,\quad $ then \begin{vmatrix} \mathbf{a} & \mathbf{\cos\alpha} & \mathbf{\sin\alpha} & \mathbf{A_X} & \mathbf{B_X} & \mathbf{B_Y} & \mathbf{A_Y} \\ \dfrac{8x+u}3 & \dfrac{10x-u}{6x} & \sqrt{1-\sin^2\alpha} & -\dfrac{13}2x & A_X+a\cos\alpha & \sqrt{R^2-B_X^2} & B_Y-a\sin\alpha \\ \dfrac{103}{19} & \dfrac{11}{38} & \dfrac{21\sqrt{3}}{38} & -\dfrac{13}2 & -\dfrac{1780}{361} & \dfrac{901}{361}\sqrt3 &-\dfrac{\sqrt3}2\\ \end{vmatrix}

The other parameters are \begin{vmatrix} \mathbf{c} & \mathbf{\cos\delta} & \mathbf{\sin\delta} & \mathbf{P} & \mathbf{D} & \mathbf{C} \\ \dfrac{2(5x+v)}3 & \dfrac{13x-v}{12x} & \sqrt{1-\cos^2\delta} & \left(\dfrac32 x,A_Y\right) & \left(\dfrac{13}2 x, A_Y\right) & D + c(-\cos\delta, \sin\delta) \\ 8 & \dfrac12 & \dfrac{\sqrt3}2 & \left(\dfrac32,-\dfrac{\sqrt3}2\right) & \left(\dfrac{13}2,-\dfrac{\sqrt3}2\right) & \left(\dfrac52,\dfrac{7\sqrt3}2\right) \\ \end{vmatrix}

One can check that $$\|A\| = \|B\| = \|C\| = \|D\| = \sqrt{43},\quad \tan\left|\dfrac{P_Y}{P_X}\right|=\dfrac{\sqrt3}3.$$

It is easy to see that the base of the first quadrilateral lies above the center of the circle, and the second - below.

(Reviving and correcting a deleted answer. Note that the first few comments refer to the previous edit.)

Ignoring reflections in $\overline{AD}$, there are two viable quadrilaterals. They're shown here as $\square AB^\prime CD$, and $\square ABC^\prime D$. Perhaps surprisingly, each shares three points and a circumcircle with the self-intersecting $\square ABCD$ that traces two equilateral triangles

$\square ABCD$ is a "near-miss" solution (that I inadvertently touted as valid in my previous edit), satisfying almost-all the requirements, with the exception that $|PB|+|PC|\neq 2|BC|$ (and, of course, the quadrilateral isn't convex). Be that as it may, the equilateral triangles make the answer to the question clear:

$$\angle OPA = 30^\circ \quad\to\quad \sin \angle OPA = \frac12$$

Currently, I don't have a straightforward way to construct $B^\prime$ and $C^\prime$, but their specifications appear below.

$$\begin{array}{c:cccc:cccc:cc} \square & |PA| & |PB| & |PC| & |PD| & |AB| & |BC| & |CD| & |DA| & \cos APB & \cos DPC \\ \hline \\ AB^\prime C D & 8 & \dfrac{157}{19} & 7 & 5 & \dfrac{103}{19} & \dfrac{145}{19} & 8 & 13 & \dfrac{4643}{5966} & \dfrac17 \\ \\ ABC^\prime D & 8 & 7 & \dfrac{199}{37} & 5 & 5 & \dfrac{229}{37} & \dfrac{256}{37} & 13 & \dfrac{11}{14} & \dfrac{829}{7363} \\ \\ \hline \end{array}$$ (In the headings, one or the other of $B$ and $C$ should be replaced by its respective counterpart, $B^\prime$ or $C^\prime$.)

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My derivation (and proof that there are no other solutions) is currently a Mathematica-aided algebraic slog. I've done my best to streamline it below. I'll note that I tried abstracting the problem a bit, using $$|AB|=p(|PA|+|PB|) \qquad |BC|=q(|PB|+|PC|) \qquad |CD|=r(|PC|+|PD|)$$ but the equations exploded into tens of thousands of terms. There may be little point in looking beyond the specific values given in the problem. (One might've expected that, if these were magic numbers, then the final edge-lengths —and especially the cosines— would be nicer. Perhaps this is an object lesson in nice-bias (hey, at least the metrics are rational), or perhaps the "intended" solution teases-out $\sin \angle OPA$ without explicitly determining the other metrics, so that their not-so-nice-ness is irrelevant. It would be interesting to see the "official" solution.)

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For the derivation, define $$a := |PA| \quad b := |PB| \quad c := |PC| \quad d := |PD|$$ $$\alpha := \angle APB \qquad \beta := \angle BPC \qquad \gamma := \angle CPD$$

where we may as well take $d=5$. The problem conditions tell us $$a = \frac85 d = 8 \qquad |AB| = \frac1{3}( b + 8 ) \qquad |BC| = \frac12(b+c) \qquad |CD| = \frac23(c+5) \tag{1}$$

By the Law of Cosines and (1), we have

$$\cos\alpha = \frac1{2ab}\left(-|AB|^2+a^2+b^2\right) = \frac{1}{18b}\left(b^2 - 2 b + 64\right) \tag{2a}$$ also $$\cos\beta = \frac{1}{8bc} \left(3 b^2 - 2 b c + 3 c^2\right) \qquad \cos\gamma = \frac{1}{18c} \left(c^2 - 8 c + 25\right) \tag{2b}$$

Now, since $\alpha+\beta+\gamma=180^\circ$, one readily shows $$1 - 2 \cos\alpha \cos\beta \cos\gamma - \cos^2\alpha - \cos^2\beta - \cos^2\gamma = 0 \tag{3}$$

Substituting from $(2)$, we (and by "we", I mean Mathematica) get ...

$$\begin{align} 0 &= 28 b^4 c^2 - 8 b^3 c^3 + 28 b^2 c^4 \\ &- 96 b^4 c - 24 b^3 c^2 - 336 b^2 c^3 - 24 b c^4 \\ &+1029 b^4 - 980 b^3 c + 1474 b^2 c^2 - 1292 b c^3 + 1497 c^4 \\ &- 600 b^3 - 12144 b^2 c - 600 b c^2 - 6144 c^3 \\ &+ 29200 b^2 - 12800 b c + 84736 c^2 \end{align} \tag{4}$$

To incorporate the cyclic nature of the target quadrilateral, I'll invoke this Lemma:

Lemma. Given $\triangle XYZ$ with circumcenter $O$ and circumradius $r$, and given some point $P$, the (signed) power of $P$ with respect to the circumcircle is given by $$p^2-r^2 = - xyz \frac{ x \sin\angle YPZ + y \sin\angle ZPX + z \sin\angle XPY }{ y z \sin\angle YPZ + z x \sin\angle ZPX + x y \sin\angle XPY } \tag{$\star$}$$ where $x := |PX|$, $y:=|PY|$, $z:=|PZ|$, $p:=|PO|$.

Proof of the Lemma is left as an exercise for the reader. (Verification using coordinates is fairly straightforward.) It's worth noting that the denominator gives a signed area of $\triangle XYZ$; I left the expression as-is so to highlight the importance of consistent angle orientation with the numerator. It's also worth noting that we could factor-out $xyz$ in the denominator, so that we have

$$p^2-r^2 = - \frac{ x \sin\angle YPZ + y \sin\angle ZPX + z \sin\angle XPY }{ \dfrac{1}{x} \sin\angle YPZ + \dfrac{1}{y} \sin\angle ZPX + \dfrac{1}{z} \sin\angle XPY }$$ which is somewhat intriguing. But I digress ...

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In our circle, we know the (signed) power of $P$ is more-conventionally given by the sub-chord product $-|PA||PD| = -40$. Applying the Lemma, we can equate this value with the power derived from, say, $\triangle ABC$, getting $$-40 = - a b c \frac{ a \sin\beta + b \sin(180^\circ+\gamma) + c \sin\alpha }{ a b \sin\alpha + b c \sin\beta + c a \sin(180^\circ+\gamma)} = - 8 b c \frac{ 8 \sin\beta - b \sin\gamma + c \sin\alpha }{ 8 b \sin\alpha + b c \sin\beta - 8 c \sin\gamma}$$ Since $\gamma = 180^\circ-\alpha-\beta$, we can manipulate above into the form $$\sin\alpha\left(-b c^2 + 40 b + c\cos\beta\left( b^2 - 40 \right)\right) = c \sin\beta \left(-3 b + \cos\alpha\left( b^2 - 40\right)\right) \tag{5}$$ Squaring both sides, rewriting sines as cosines, and substituting from $(2)$, "we" find

$$0 = (b + c) (b + 8) \left(\begin{align} &\phantom{+} 4 b^4 c^2 - 80 b^3 c^2 \\ &- 6 b^4 + 340 b^3 c + 250 b^2 c^2 \\ &- 2310 b^3 + 1300 b^2 c - 2310 b c^2 \\ &+ 9175 b^2 + 11230 b c + 2775 c^2 \\ &- 128000 b + 409600 \end{align}\right) \tag{6}$$

From here, Mathematica finds that the system $(4)$ and $(6)$ has many non-real solutions for $b$ and $c$, yet only three cases where both values are real: the extraneous case $b=c=0$, and the values shown in the table above. This completes the derivation. $\square$

Let $AD=x$, $PB=y$, and $PC=z$, than we find from the four constraints that \begin{eqnarray} AP & = & \frac{8}{13}x\\ DP & = & \frac{5}{13}x\\ AB & = & \frac{8x+13y}{39}\\ BC & = & \frac{y+z}{2}\\ CD & = & \frac{2(5x+13z)}{39} \end{eqnarray} For the angle at $A$, we have that $$ \cos(\angle BAP) = \frac{AB^2+AP^2-BP^2}{2 AB ~ AP} = \frac{AB^2+AD^2-BC^2-CD^2}{2 (AB ~ AD + BC ~ CD)} $$ where the first term is using the cosine rule for the triangle $BAP$ and the second expression is a similar result for the cyclic quadrilateral $ABCD$. Substitution of the results above we obtain an equation in $x,y,z$: $$f(x,y,z)= -\frac{1780 x^3-3120 x^2 y-4680 x^2 z+11323 x y^2-6422 x y z-10985 x z^2+8788 y^2 z+8788 y z^2}{312 x \left(8 x^2+18 x y+5 x z+13 y z+13 z^2\right)}=0. $$ In a similar fashion, we can derive another equation for the angle at $C$ by using the triangle $CDP$ and the quadrilateral $ABCD$: $$g(x,y,z)= -\frac{2848 x^3-2184 x^2 y-4680 x^2 z-4394 x y^2-3887 x y z+11323 x z^2+2197 y^2 z+2197 y z^2}{156 x \left(20 x^2+8 x y+60 x z+13 y^2+13 y z\right)} = 0. $$ Since $x,y,z>0$ we can focus on the numerators only, which are second order in $z$. The coefficients of the term $z^2$ are $2197(5 x -4y)$ and $-169(67x+13y)$ respectively. Note that both these factors are non-zero, this is obvious for the second, whereas for the first is follows from the fact that in the triangle $ABP$ we have $AB+AP>BP$. This gives $16 x > 13 y$ and hence $5 x > 4 y$. We can therefore combine the numerators of both expressions to eliminate the term proportional to $z^2$, which results in a linear equation in $z$ and the solution: $$ z = \frac{33820 x^3-52884 x^2 y+60671 x y^2+41743 y^3}{13 \left(5280 x^2+4277 x y-7267 y^2\right)}. $$ If we substitute this back in the numerators of $f(x,y,z)$ or $g(x,y,z)$ we obtain $$ \frac{52 (157 x-247 y) (7 x-13 y) (70 x-13 y) (5 x-4 y) (50 x+39 y) \left(89 x^2+78 x y+338 y^2\right)}{\left(5280 x^2+4277 x y-7267 y^2\right)^2} = 0 $$ or $$ -\frac{4 (157 x-247 y) (7 x-13 y) (70 x-13 y) (67 x+13 y) (50 x+39 y) \left(89 x^2+78 x y+338 y^2\right)}{\left(5280 x^2+4277 x y-7267 y^2\right)^2}=0. $$ Since they still have to be satisfied both and $x,y>0$ there are only three possible solutions, and in combination with the results for $z$ we find the following results: \begin{eqnarray} & y=\frac{157}{247}x \qquad & z=\frac{7}{13}x\\ & y=\frac{7}{13}x \qquad & z=\frac{199}{481}x\\ & y=\frac{70}{13}x \qquad & z=-\frac{44}{13}x \end{eqnarray} and hence we have to discard the third solution.

Let $O=(0,0)$ be the centre of the circle, and choose $A=(-\frac{x}{2},h)$ and $B=(\frac{x}{2},h)$ on the circle of unit radius. For the point $P$ we easily find $P=(\frac{3x}{26},h)$. It is then straightforward to obtain the coordinates for $B$ and $C$ by \begin{eqnarray} B & = & P + y (-\cos \angle APB,\sin \angle APB)\\ C & = & P + z ( \cos \angle CPD,\sin \angle CPD) \end{eqnarray} With the help of the above results and the equations for the distances $OA=OB=OC=OD=1$, we can find the correct values for $x$ and $h$. Doing so, results in two distinct solutions $$ (x,y,z,h) = (\frac{13}{\sqrt{43}},\frac{157}{19 \sqrt{43}},\frac{7}{\sqrt{43}},-\frac{\sqrt{3}}{2\sqrt{43}}) $$ $$ (x,y,z,h) = (\frac{13}{\sqrt{43}},\frac{7}{\sqrt{43}},\frac{199}{37 \sqrt{43}},\frac{\sqrt{3}}{2\sqrt{43}}) $$

In either case one finds that $\sin \angle OPA = \frac{1}{2}$.

It is quite unsatisfactory to solve a geometry problem by means of analysis, but perhaps with knowledge of the answer someone can find a more elegant solution.