$\bigtriangleup{AMP} \sim \bigtriangleup{ANQ}$.

Let $T$ be the intersection of $PM$ and $QN$. We want to prove that $A, M, N, T$ are concyclic, which is the same as $\angle AMT = \angle ANT$, which in turn is equivalent as $\angle AMP = \angle ANQ$. We will prove something stronger: Triangles $AMP$ and $ANQ$ are similar. For this, let $I$ be the incenter of $ABC$. Since $BN$ and $CM$ are internal angle bisectors of angles $ABC$ and $ACB$, $I$ is in both of them. It is known that $M$ and $N$ are the respective circumcenters of triangles $AIB$ and $AIC$, which implies $AM = MI$ and $AN = NI$. It follows that triangles $AMN$ and $IMN$ are similar. Next, let us observe that $\angle IMN = \angle IBC$ and $\angle INM = \angle ICB$, because $B, C, M, N$ are concyclic. From this is follows that $IMN$ and $IBC$ are similar. Finally, observe that $$\angle BAP = \frac{180^{\circ} - \angle ABP}{2} = \frac{\angle ABC}{2} = \angle ABI$$ Then $AP$ is parallel to $BI$. Analogously $AQ$ is parallel to $CI$. This implies that triangles $APQ$ and $IBC$ are homothetic and thus simlar. From the similarities obtained before we get that $AMN$ and $APQ$ are similar, which implies $$\angle PAM = \angle CAN \hspace{2em} \text{ y } \hspace{2em} \frac{AP}{AM} = \frac{AQ}{AN}$$ So we get that triangles $AMP$ and $ANQ$ are similar, as desired.

It is easy to see that $\triangle APQ \sim \triangle AMN$ (it is trivial angle chase ). Thus there exists spiral similarity sending $MN$ to $PQ$ centered at $A$. Consequently $PM \cap QM \in \odot(AMN)=\odot(ABC)$ and $PM \cap QN \in \odot(APQ)$ as desired.