WLOG $P$ and $B$ are on opposite sides of $AC$. Let $I$ be the incentre of $ABC$. Note that $AMIN$ cyclic, so $\angle MAI = \angle MNI$. Further note that $\angle INC = 90^{\circ}$, so $\angle PNC = 90^{\circ} - \angle MAI$. However, $\angle MAI = \frac{1}{2} \angle A = 90^{\circ} - \frac{1}{2} (\angle B + \angle C) = 90^{\circ} - (\angle IBC + \angle ICB) = 90^{\circ} - \angle PIC$, so $\angle PIC = \angle PNC$, so $PNIC$ is cyclic. Thus, $\angle BPC = \angle IPC = \angle INC = 90^{\circ}$.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/An%20unlikely%20concurrence.pdf p. 5-6. Sincerely Jean-Louis