Let $I$ be the incentre of the triangle $ABC$.

Triangle AMC is isosceles with the apex at A, so the line AP is the altitude to the base and at the same time the bisector of the angle $\angle MAC$. It follows that $I$ lies on $AP$. Similarly, $I$ lies on the line $BR$. This implies that $\angle CPI = \angle CPA = \pi/2$ and $\angle IRC = \angle BRC = \pi/2$, so the points $P$ and $R$ lie on the circle with diameter $CI$. Since $D$ and $E$ are the points where the incircle touches the sides of the triangle, we have $\angle CDI = \angle IEC = \pi/2$, and $D$ and $E$ therefore lie on the circle with diameter $CI$. We have shown that the points $P$, $R$, $D$ and $E$ are concyclic.