The bisector of angle $\angle EBA$ and the perpendicular bisector of side $EA$, both bisect the $\operatorname{arc}\overparen{EA}$ of the circumcircle of triangle $ABE$ opposite to the vertex $B$. Therefore quadrilateral $ABEM$ is cyclic and hence
$$\angle EAM = \angle EBM$$
The segment defined by the two intersection points of the perpendicular bisector of $AE$ with the circumcircle of triangle $AED$ is a diameter and intersects the bisector of the angle $\angle ADE$ on this circumcircle. But the bisectors of angles $\angle CDA$ and $\angle ADE$ are perpendicular. Therefore quadrilateral $AEDN$ is cyclic and hence
$$\angle CDN = \angle EAN.$$

We deduce that
$$\angle MAN = \angle EAN - \angle EAM = \angle CDN - \angle EBM = \angle MPN,$$
from triangle $BDP$. This proves that points $A, M, N, P$ are concyclic.