Square roots and cube roots equation

Let $x:=\sqrt{p}$ so $0=1-x^2-(1-x)^3=x(1-x)(3-x)$ and$$x\in\{0,\,1,\,3\}\implies p=x^2\in\{0,\,1,\,9\}\implies a\in\left\{\tfrac43,\,\tfrac53,\,\tfrac{13}{3}\right\},$$all of which work.

Using your first substitution:

$$\begin{align*} \sqrt p+\sqrt[3]{1-p}&=1\\[1ex] \sqrt[3]{1-p}&=1-\sqrt p\\[1ex] 1-p&=(1-\sqrt p)^3\\[1ex] 1-p&=1-3\sqrt p+3p-|p|\sqrt p\\[1ex] \end{align*}$$

If $p=3a-4\ge0$ (and this is the only case if you're looking for real-valued solutions, since $\sqrt p$ would be defined only for $p\ge0$), or $a\ge\frac43$, then $|p|=p$:

$$\begin{align*} 1-p&=1-3\sqrt p+3p-p\sqrt p\\[1ex] 1-p&=1+3p-(3+p)\sqrt p\\[1ex] 4p&=(3+p)\sqrt p\\[1ex] \frac{4p}{3+p}&=\sqrt p\\[1ex] \frac{16p^2}{(3+p)^2}&=p\\[1ex] 16p^2&=p(3+p)^2\\[1ex] 16p^2&=p^3+6p^2+9p\\[1ex] p(p-1)(p-9)&=0 \end{align*}$$

Then either $p=0$, $p=1$, or $p=9$, which means either $a=\frac43$, $a=\frac53$, or $a=\frac{13}3$.

It is indeed useful. Write $$ \sqrt[3]{1-p}=1-\sqrt{p} $$ and cube: $$ 1-p=1-3\sqrt{p}+3p-p\sqrt{p} $$ Writing $r=\sqrt{p}$ you get $$ r^3-4r^2+3r=0 $$ and the rest should be easy. Just beware that the solutions are subject to $r\ge0$ and $p\ge0$ (in this case there's no problem, however).