Solution:

$$\begin{aligned} \angle QCP &= \angle QCA + \angle ACN + \angle NCP \\ &= \angle CNA + \angle ACN + \angle NAC \\ &= 180^\circ, \end{aligned}$$

i.e. the points $P$, $C$ and $Q$ are collinear.

b) Set $\angle ACM = \varphi$, $\angle NCM = \psi$ and denote by $R_1$ and $R_2$ the radii of the circumcircles of $\triangle AMC$ and $\triangle MNC$, respectively. Then by the Sine theorem we have
$$\begin{aligned} QC &= 2R_1 \sin \angle QMC = 2R_1 \sin \psi \\ CP &= 2R_2 \sin \angle PMC = 2R_2 \sin \varphi \\ AM &= 2R_1 \sin \varphi \\ BM &= 2R_2 \sin \psi \end{aligned}$$
Therefore $PC \cdot CQ = 4R_1 R_2 \sin \varphi \sin \psi = AM \cdot BM$.

As in a) we see that the points $L$, $C$ and $K$ are collinear. Analogously we have $CK \cdot CL = AN \cdot BN$. Since the circumcircle of $\triangle MNC$ passes through $K$, $C$ and $P$, the lines $PQ$ and $KL$ do not coincide. Therefore the points $P$, $Q$, $K$ and $L$ are cyclic if and only if
$$CP \cdot CQ = CL \cdot CK \Longleftrightarrow AM \cdot BM = AN \cdot BN \Longleftrightarrow AM = BN.$$