Note that $\triangle{PAD}$ is equilateral, assume that coordinate of $A,B,C,D$ are $A(0,0),B(0,1),C(1,1),D(1,0)$ We get $P(\frac{1}{2},\frac{\sqrt{3}}{2})$ and $Q$ lie on the line $y=\frac{1-\frac{\sqrt{3}}{2}}{1-\frac{1}{2}} x$ and circle center at $C(1,1)$ with radius $PA=AD=1$ So we get $Q(\frac{2+\sqrt{3}}{2},\frac{1}{2})\rightarrow N(\frac{3+\sqrt{3}}{4},\frac{\sqrt{3}+1}{4})$, so $tan(\angle{NAD})=\frac{\frac{\sqrt{3}+1}{4}}{\frac{3+\sqrt{3}}{4}} =\frac{1}{\sqrt{3}}$ We get $\angle{NAD}=30^{\circ}\rightarrow \angle{CAD}=15^{\circ}$ and we easily prove that $\angle{CNA}=90^{\circ}$ Finally we get $\angle{CAD}=15^{\circ},\angle{CNA}=90^{\circ},\angle{NCA}=180-15-90=75^{\circ}$

Solution: Let $BD\cap AC = M$. Join $MN$ and $DQ$. Let $a$ be the length of the side of the square. Simple length and angle chasing proves that $D$ is the circumcenter of cyclic $APCQ$. This gives $AB = BC = CD = DA = DQ = CQ = AP = a$. Since, $AP = AB$, so, $\angle APB$ = $75$°. Also, $\angle APQ$ = $105$°. Thus, $B$, $P$ and $Q$ are collinear. So, $\angle QBC$ = $15$°. Now, $MN || AQ$. So, $\angle MNB$ = $30$° = $\angle MBN$. This gives $MN = MB = MC = MD = MA$. Thus, $ADNCB$ is cyclic. Trivial angle chasing gives $\angle ANC$ = $90$°, $\angle NAC$ = $15$° and $\angle NCA$ = $75$°.