Let $\left(M \right)$ be the circumcircle of triangle $ABC,$ $ \ J = NM \cap \left(M \right), \ K = CM \cap \left(M \right).$ We know that $JN$ perpendiculars to $BC$ and $\angle AMJ = \angle JMK, \ \angle JMC = \angle JMB, $ so $\Delta JMC = \Delta JMB \implies \angle JCM = \angle JBM = \angle JCA.$ So $J$ is the $C- \text{excircle}.$ Now $J \in \left(ABC \right), JN,JQ,JP$ are perpendicular $BC,AB,AC,$ so $N,P,Q$ are collinear.

easy by Menelaus' theorem