Nice problem!

I only found not-quite short bash solution,here is sketch: $\frac{[APC]}{[APQ]}=\frac{b}{AQ}$ and $\frac{[CPB]}{ACP}=\frac{PB}{AP}$ so we have $[CPB]=\frac{b\cdot PB}{AQ\cdot AP}\cdot [APQ]$ so from $[APQ]=[BCQP]$ we have $AP\cdot CQ+b\cdot PB=AQ\cdot AP$ and now from from $P-O-B$ applying in bary formula for colinearity we have: $\frac{QC(AP\cdot \sin(2C))}{\sin(2A)}+AQ\cdot PB\cdot \frac{\sin(2B)}{\sin(2A)}=AQ\cdot AP$ so we have to prove these to are equivalent,ie. $LHS$ of these to are equal.

This problem was proposed by Burii. The assumption that $D$ is the feet of angle bisector is redundant. It can be any point on the side $BC$ as can be clearly seen in the following solution. Let $AX$ be a diameter of the circumcircle of $ABC$. Observe that $BX \parallel DP$, so $[BDP]=[XDP]$. Analogously, $[CDQ]=[XDQ]$. Thus $$[BCQP]=[BDP]+[PDQ]+[CDQ]=[XDP]+[BDP]+[XDQ]=[PXQ].$$Therefore $[APQ]=[BCQP] \iff [APQ]=[PQX] \iff O \in PQ$, where $O$ is the midpoint of $AX$, i.e. the circumcenter of $ABC$.

Notice that $[APQ]=[BCQP] \iff [APQ]=\frac{[ABC]}{2} \iff \frac{AP \cdot AQ \cdot \sin \angle BAC}{2}=\frac{AB \cdot AC \cdot \sin \angle BAC}{4} \iff AP \cdot AQ=\frac{AB \cdot AC}{2}$. Consider transformation $\phi$- composition of symetry wtr bisector of $\angle BAC$ and inversion in $A$ with radius $\sqrt{\frac{1}{2} \cdot AB \cdot AC}=AP=AQ$. Denote $M, N$ as midpoints of $AB, AC$ and let $H$ be a feet of $A$-altitude. It is sufficient to show that $A, \ \phi(O), \ \phi(P)=Q, \phi(Q)=P$ are concyclic. Notice that $\phi(B)=N, \ \phi(C)=M$. Since $B, H, C$ are collinear, $A, M, \phi(H), N$ are concyclic. However, as $AH$ and $AO$ are isogonals wtr $\angle BAC$ and $\angle CHA=\angle BHA=\frac{\pi}{2}=\angle AN\phi(H)=\angle AM\phi(H)$, hence $\phi(H)=O$ and $\phi(O)=H$. Points $A, P, H, D, Q$ are concyclic, so we are done.