Solution:

Note that $N$ is also the midpoint of $PS$. From right-angled triangles $PAS$ and $CQR$ we obtain $\angle ANP = 2 \angle ASP$, $\angle CNQ = 2 \angle CRQ$, hence
$$\angle ANC = \angle ANP + \angle CNQ = 2(\angle ASP + \angle CRQ) = 2(\angle RSD + \angle DRS) = 2 \angle ADC.$$
Similarly, using right-angled triangles $BAD$ and $BCD$, we obtain $\angle AMC = 2 \angle ADC$.
Thus $\angle AMC = \angle ANC$, and the required statement follows.

Solution:

In this proof we show that we have $\angle NCM = \angle NAM$ instead. From right-angled triangles $BCD$ and $QCR$ we get $\angle DRS = \angle CRQ = \angle RCN$ and $\angle BDC = \angle DCM$. Hence $\angle NCM = \angle DCM - \angle RCN$. From right-angled triangle $APS$ we get $\angle PSA = \angle SAN$. From right-angled triangle $BAD$ we have $\angle MAD = \angle BDA$. Moreover, $\angle BDA = \angle DRS + \angle RSD - \angle RDB$.
Therefore $\angle NAM = \angle NAS - \angle MAD = \angle CDB - \angle DRS = \angle NCM$, and the required statement follows.

Solution:

As $N$ is also the midpoint of $PS$, we can shrink triangle $APS$ to a triangle $A_0 QR$ (where $P$ is sent to $Q$ and $S$ is sent to $R$). Then $A_0, Q, R$ and $C$ lie on a circle with center $N$. According to the shrinking, the line $A_0R$ is parallel to the line $AD$. Therefore $\angle CNA = \angle CNA_0 = 2 \angle CRA_0 = 2 \angle CDA = \angle CMA$. The required statement follows.