Lemma: let $ABCDE$ be a cyclic pentagon with circumcenter $O$ and $M$,$N$,$P$ and $Q$ be the midpoints of $AE$,$AB$,$BC$ and $CD$. let $P'$ be the reflection of $P$ through midpoint of $NQ$ then $ANP'OM$ is cyclic. proof: note that $A$, $P'$ and $D$ are collinear so, \[\angle AMN=\angle AEB=\angle ADB=\angle AP'N\]as desired. now consider that $M$,$N$,$P$ and $Q$ are the given points in the problem in that order. $P'$ and $N'$ are the reflections of $P$ and $N$ through midpoints of $NQ$ and $PM$. by the lemma we can find the circumcenter $O$ since it is the intersection of $\odot MNP'$ and $\odot PQN'$. now we have the circumcenter so we can instantly find the pentagon and we are don. $\blacksquare$