Let $AC,BD$ intersect at $E$. By Miquel's, circumcircles of triangle $AMP,DMQ,EPQ,EAD$ meet at a common point, circumcircles of $BNQ,CNP,EPQ,EBC$ meet at a point. Hence it remains to show that circumcircles of $EPQ,EAD,EBC$ meet at a point. Let circumcircles of $EBC,EAD$ meet at $F$. Hence $F$ is the spiral center sending $BD$ to $AC$. Since $\frac{AP}{AC}=\frac{DQ}{BD}$, hence $FDB$ similar to $FAC\Rightarrow FPA$ similar to $FQD$. As the center of spiral similarity sending $PA$ to $QD$ is unique, hence $F$ lies on circumcircle of $EPQ$.