Again this seems easy as well. Hope I am not wrong. Let $AC\cap BD=E$. By Miquel's, $\odot AMP, \odot DMQ, \odot EPQ, \odot EAD$ concur, and $\odot BNQ, \odot CNP,\odot EPQ, \odot EBC$ concur as well. Hence it remains to show that $\odot EPQ,\odot EAD,\odot EBC$ concur. Let $\odot EBC\cap \odot EAD=F$. Hence $F$ is the center of the spiral similarity sending $BD$ to $AC$. Since $\frac{AP}{AC}=\frac{DQ}{BD}$, hence $\Delta FDB\sim \Delta FAC\implies \Delta FPA\sim \Delta FQD$. As the center of spiral similarity sending $PA$ to $QD$ is unique, hence $F$ lies on $\odot EPQ$. Thus our proof is complete. Note that we don't need $P,Q$ to be midpoints, just $\frac{AP}{AC}=\frac{DQ}{BD}$.

I shall prove a stronger statement: In a quadrilateral $ABCD$ the diagonals $AC$ and $BD$ meet at $K$.Let $P$ and $Q$ be the midpoints of $AC$ and $BD$ respectively.Let the line $PQ$ meet $AB$ at $N$ and $CD$ at $M$.Then the six circles $APN,BQN,CPM,DQM,KQP,DKC$ have a common point. Let $J$ be the intersection point of the circles $APN$ and $BQN$.Then note that $\angle{PAJ}=\angle{PNJ}=\angle{JNQ}=\angle{JBQ}$ and $\angle{PJA}=\angle{PNA}=\angle{QNB}=\angle{QJB}$.Thus $\triangle{APJ} \sim \triangle{BQJ}$.Since these triangles are the triangles formed by the $J$ medians of $\triangle{CQA}$ and $\triangle{DJB}$ we have $\triangle{BDJ} \sim \triangle{ACJ}$.Hence we have $\angle{KDJ}=\angle{BDJ}=\angle{ACJ}=\angle{KCJ}\implies DCJK$ is cyclic.Also see that $\angle{KQJ}=180-\angle{BQJ}=\angle{JNB}=\angle{JNA}=\angle{JPK} \implies KJQP$ is also cyclic.Using these facts we obtain $\angle{MCJ}=\angle{DCJ}=\angle{QKJ}=\angle{QPJ}=180-\angle{JPM}$ so $MPJC$ is cyclic.Also $\angle{JQD}=\angle{JQK}=\angle{JPC}=\angle{JMD}$ so $JQMD$ is cyclic.Finally $\odot{APN},\odot{BQN},\odot{CPM},\odot{QDM},\odot{KQP},\odot{DKC}$ concur at $J$.

This problem is just a special case of 2006 USAMO Problem 6