Let $A'$ be the reflection of $A$ in the midpoint of $BC$ and $T$ the intersection of tangents to $(ABC)$ at $B$ and $C$. Note that $ABA'C$ is a parallelogram and thus $A'C=AB=AP$. Combining this with the fact that $PB\parallel AC$, we get that $APA'C$ is an isosceles trapezoid. Now we see that the perpendicular bisector of $PA'$ is also the perpendicular bisector of $AC$, i.e. $OP=OA'$. Similarly, $OQ=OA'$ so we conclude that $O$ lies on the perpendicular bisector of $PQ$. By DDIT on $\mathcal{Q}\{BP,PC,CQ,QB\}$ we get an involution that swaps $(AB,AC)$, $(AP,AQ)$, $(AX,AA')$. Note that $\angle PAB=180^{\circ}-2\angle A=\angle CAQ$ so $AP$ and $AQ$ are reflections of each other in the $A$-angle bisector. This implies that $AX$ and $AA'$ are reflections in the $A$-angle bisector. But since $AA'$ is the median, $AX$ must be the symmedian, i.e. $A$, $X$ and $T$ are colinear. Keeping in mind that $APA'C$ is a cyclic isosceles trapezoid, $\angle BPX=\angle A'PC=\angle A'AC=\angle BAX$ so $APBX$ is cylic. Now we have $\angle BXT=\angle BPA=\angle A=\angle BOT$ so $X\in(BOCT)$. Note that $\angle PXA=\angle A=\angle QXA$ so $XA$ is the internal angle bisector of $\angle PAQ$. But sine $X\in(BOCT)$ we get $XO\perp AX$ so combining this with the fact that $O$ lies on the perpendicular bisector of $PQ$, we recognize $O$ as the midpoint of arc $PXQ$ in $(PXQ)$ so the problem is done.

Attachments:

Note that $\triangle APB\sim\triangle ACQ$, so there is a spiral similarity sending $PB\to CQ$ centered at $A$. In particular, this implies $X$ lies on $(APB)$ and $(ACQ)$ by phantom points. This means that $\angle PXA=\angle PBA=\angle BAC$, so by symmetry, $\angle PXA=\angle AXQ=\angle BAC$. As $\angle BXC=\angle BOC=2\angle BAC$, so $X\in (BOC)$. As $O$ is the midpoint of arc $BC$, we know $XO$ is the exterior angle bisector of $\angle BXC$ or $\angle PXQ$. It thus suffices to show that $OP=OQ$, as then $O$ is the midpoint of arc $PQ$ in $(PQX)$. This is simple as if $A'$ is the intersection of $BP$ and $CQ$, then as $ABA'C$ is a parallelogram, we know $A'P\parallel AC$ and $AP=AB=A'C$. Thus $A'P$ and $AC$ have the same circumcenter, so by symmetry, $O$ is the circumcenter of $A'PQ$.

Here is a clean complex bash: $$p=\frac{ab+ac-bc}{b}$$$$q=\frac{ab+ac-bc}{c}$$($\triangle PAB \sim \triangle CTB$ where $T=\frac{2bc}{b+c}$) $p,q,o,x$ are concylcic iff $$\vert \frac{p-c}{q-b}\vert = \vert \frac{p}{q}\vert$$Which is true since both equal $\vert \frac{c}{b}\vert$