We denote by $Q$ the intersection of $CZ$ and $AB$. First, we show that quadrilateral $QXYP$ is cyclic. Note that $\angle PNA = \angle PYQ$. Therefore, if we prove that $\angle QXP = \angle PNA$, it follows that $QXYP$ is cyclic. This is also equivalent to proving that $XQ$ is parallel to $AN$.

To prove this, we use $R$, which is the intersection of $AZ$ and $BC$. We can write:
$$\angle RAP = \angle PNA = \angle PMN = \angle PBA = \angle PBR$$
Thus, $R$ also lies on the circumcircle of triangle $APB$. Clearly, $AC \parallel QR$, so we only need to show that $RQ$ passes through $X$. Note that:
$$\angle QRA = \angle RAN = \angle XPA = \angle XRA$$
This equation implies that $R$, $Q$, and $X$ are collinear, yielding $QXYP$ is cyclic. Note that $C$, $N$, and $Q$ are reflections of $A$, $M$, and $R$ with respect to line $BP$. Moreover, since the pentagon $BRPMA$ is cyclic, we can conclude that:
$$\angle CQP = \angle PNA = \angle PXQ$$
Implies that $CZ$ is tangent to the circumcircle of $PXY$. $\blacksquare$