Let $AD$ intersect $BC$ at $E$. Note that $ABCD$ is cyclic and $\angle AEC=20$. Since $\triangle EDC\sim\triangle EBA$, we have $\triangle EDN\sim\triangle EBM$, so $\frac{EN}{EM}=\frac{ND}{MB}=\frac{NP}{PM}$ so $EP$ is the angle bisector of $\angle NEM$. But since $\angle DEN=\angle BEM$, $EP$ is also the angle bisector of $\angle AEC$. But since $AP=PC$ and $EA\neq EC$(hopefully), we have $EAPC$ is cyclic, so $\angle APC=180-20=160$.