Bounding the non-multiplicativity of isometric projection

If the constant is allowed to depend upon the dimension, the estimate is simple. Let $A=O_AP_A,B=O_BP_B$, then $AB=O_AO_B(O_B^*P_AO_B)P_B$ and we are left with showing that if the product of two positive definite self-adjoint operators $X=O_B^*P_AO_B$ and $Y=P_B$ is $\delta$-close to a unitary operator $V$, then it is $C\delta$-close to $I$. Now, we do not really know much about $XY$ except that it is conjugate to $X^{1/2}YX^{1/2}$, so all eigenvalues are real positive. Thus, it will suffice to show that if a $\delta$-perturbation of a unitary operator $U$ has real positive eigenvalues, then $U$ is $C\delta$-close to the identity. However, if it were not the case, there would be an eigenvalue of $U$ that is $\frac 12 C\delta$ far from the positive semi-axis. If you now take the Gershgorin disks of radius $\delta$ and if $C>5n$, say (where $n$ is the matrix size), then there would be a connected component of the union of the Gershgorin disks that would not cross the positive semi-axis, so some eigenvalues would be confined there.

I'm still curious if we can get a dimension-independent bound, so don't hurry to accept this answer ;-)

Edit. Since there are some difficulties in understanding, let me go into some details (all of which are totally classical).

Gershgorin-Rouche theorem Let $A_t\quad (t\in[0,1])$ be a continuous family of matrices and $K$ a compact set on the complex plane (with decent boundary, if you want, though it is irrelevant). if the boundary of $K$ is free from the eigenvalues of $A_t$ for all $t\in(0,1)$, then all $A_t$ have the same number of eigenvalues in $K$. In particular, if $A_0$ is normal and $A_t=A_0+tQ$, then each connected component $K$ of the union of closed disks of radius $\|Q\|$ centered at the eigenvalues of $A_0$ has exactly as many eigenvalues of $A_1$ in it as of $A_0$.

Proof The first part is just the classical Rouche theorem about zeroes of analytic functions applied to the characteristic polynomials of $A_t$. To show that second part, let us just show that no $A_t$ can have an eigenvalue $\lambda$ on the boundary of $K$. Indeed, for any non-zero vector $x$, we have $|(A_0-\lambda I)x|\ge \|Q\||x|$ (here we use the normality of $A_0$) and $|tQx|<\|q\||x|$, so, by the triangle inequality $|(A_t-\lambda I)x|>0$.

Now, take $A_0=U$ here and let $K$ be the connected component of the union of disks of radius $\delta$ containing the "faraway" (from the positive real semi-axis) eigenvalue of $U$. Then any $\delta$-perturbation of $U$ must have eigenvalues in $K$. Note that those eigenvalues do not need to be close to any particular eigenvalue of $U$, but they are certainly not real positive, and that's all I need to get a contradiction.