, 2021

Since $P$ is the midpoint of the minor arc $BC$, $AD$ is the internal angle bisector of $\angle BAC$.
Let $E'$ be the intersection of the perpendicular bisector of $AD$ and the line through $A$ perpendicular to $AB$. Using the method of false position, it suffices to show $BE' \perp MP$, and then conclude $E' = E$. Let $\Omega$ be the circle with centre $E'$ that passes through $A$ (and hence $D$).

Since $AB \perp AE'$, $AB$ is tangent to $\Omega$. As $MA = MB$, the powers of $M$ with respect to $\Omega$ and $B$ are equal. Since $\angle PBD = \angle PAC = \angle PAB$, we have $\triangle PBD \sim \triangle PAB$, so $PB^2 = PD \times PA$. Therefore, the powers of $P$ with respect to $\Omega$ and $B$ are equal.
From above, both $M$ and $P$ lie on the radical axis of $\Omega$ and $B$. Hence, $MP$ is perpendicular to the line joining the centres of the two circles, i.e. $E'B$. The result follows from above.