Solution:

Refer to the figure shown below.

Since $C P$ is the $C$-altitude of $A B C$, triangles $A C P$, $A B C$ and $C B P$ are similar, so $A P = \frac{A C^{2}}{A B}$ and $B P = \frac{B C^{2}}{A B}$. As $C N Q B$ is cyclic, we have $\angle P N Q = \angle C B P = \angle A C P$ so triangles $A C P$ and $Q N P$ are similar. With $N$ being the midpoint of $P C$, we have $\frac{A P}{P Q} = \frac{P C}{P N} = 2$, which gives $P Q = \frac{1}{2} A P$ and $B Q = A P - A Q = A B - \frac{3}{2} A P$.

We now claim that $S C^{2} = 2 A B \cdot P M$. Indeed, since $Q R$ and $C P$ are parallel, we have $\frac{B Q}{B P} = \frac{B R}{B C}$ by Thales' Theorem. Thus,
$$R S = B R = \frac{B C \cdot B Q}{B P} = \frac{B C\left(A B - \frac{3 A C^{2}}{2 A B}\right)}{\frac{B C^{2}}{A B}} = \frac{2 A B^{2} - 3 A C^{2}}{2 B C}$$
Applying Pythagorean theorem on triangle $S C R$, we see that
$$\begin{aligned} S C^{2} & = B R^{2} - R C^{2} = B C (2 B R - B C) = B C\left(\frac{2 A B^{2} - 3 A C^{2}}{B C} - B C\right) \\ & = 2 B C^{2} + 2 C A^{2} - 3 C A^{2} - B C^{2} = B C^{2} - C A^{2} \end{aligned}$$
Note that $\tan A > \sqrt{2}$ implies that $S$ is indeed on ray $C A$ past $A$. On the other hand, with $M$ being the midpoint of $A B$, we compute
$$P M = A M - A P = \frac{A B}{2} - \frac{C A^{2}}{A B} = \frac{A B^{2} - 2 C A^{2}}{2 A B} = \frac{B C^{2} - C A^{2}}{2 A B} = \frac{S C^{2}}{2 A B}$$
which proves the desired claim.

Observe that the circumcircle of $C P B$ is tangent to $S C$ at $C$, so the Power of the Point gives $S C^{2} = S P \cdot S T$. It follows from the above claim that
$$S P \cdot S T = 2 A B \cdot P M \Longleftrightarrow \frac{S P}{P M} = \frac{2 A B}{S T} = \frac{2 A B}{2 A W} = \frac{A B}{A W}$$
Since $A V = P V$, we have $\angle W A B = 180^{\circ} - \angle V A P = 180^{\circ} - \angle V P A = \angle M P S$, so triangles $W A B$ and $M P S$ are similar. Thus, we see that $\angle P S M = \angle A B W$ and with $O P = O M$, we arrive at
$$\angle A B W + \angle P M O = \angle P S M + \angle P M O = \frac{1}{2}(\angle P O M + 2 \angle P M O) = \frac{1}{2} \cdot 180^{\circ} = 90^{\circ}$$
Hence, we get $O M \perp B W$, which completes the proof.