Prove some points are concyclic

Note that $\angle APB = \angle CNA$ and $\angle BAP = \angle NAC$. This shows that $\triangle APB \sim \triangle ANC$. Similarly, $\angle AMB = \angle CQA$ and $\angle BAM = \angle QAC$, hence $\triangle AMB \sim \triangle AQC$.

We have $$\angle PMN = \angle PMA = \angle PBA = \angle ACN = \angle AQN = \angle PQN,$$ hence $P,Q,M,N$ are concyclic.

Let $BM$ intersect the circumcircle of $PQMN$ again at $T\neq M$. Note that $\angle PTM = \angle PQM$ and $\angle MAP = \angle MBP$, hence $\triangle BPT \sim \triangle AMQ$. Therefore $\dfrac{BT}{AQ}=\dfrac{BP}{AM}$. On the other hand, similarity $\triangle AMB \sim \triangle AQC$ gives $\dfrac{BM}{AM}=\dfrac{CQ}{AQ}$. Hence the power of $B$ with respect to the circumcircle of $PQMN$ is equal to $$BM \cdot BT = \frac{AM\cdot CQ}{AQ} \cdot \frac{BP \cdot AQ}{AM} = CQ \cdot BP.$$

Similarly, let $CQ$ intersect the circumcircle of $PQMN$ again at $U\neq Q$. We have $\angle APN = \angle CUN$ and $\angle NAP = \angle NCU$, hence $\triangle CUN \sim \triangle APN$ which gives $\dfrac{CU}{CN}=\dfrac{AP}{AN}$. Moreover, $\triangle APB \sim \triangle ANC$ gives $\dfrac{CN}{AN}=\dfrac{BP}{AP}$. Therefore the power of $C$ with respect to the circumcircle of $PQMN$ is equal to $$CQ\cdot CU = CQ \cdot \frac{CN\cdot AP}{AN} = CQ \cdot \frac{\frac{BP\cdot AN}{AP}\cdot AP}{AN} = CQ \cdot BP.$$

This shows that $B$ and $C$ have equal powers with respect to the circumcircle of $PQMN$. Denote the center and the radius of this circle by $O$ and $r$. Then $BO^2-r^2=CO^2-r^2$ from which it follows that $BO=CO$. In other words, $O$ lies on the perpendicular bisector of $BC$.