Solution:

Extend $CP$ to $D$ such that $CP = PD$. Let $\angle BCP = \alpha = \angle BAP$. Observe that $BP$ is the perpendicular bisector of $CD$. Hence $BC = BD$ and $BCD$ is an isosceles triangle. Thus $\angle BDP = \alpha$. But then $\angle BDP = \alpha = \angle BAP$. This implies that $B, P, A, D$ all lie on a circle. In turn, we conclude that $\angle DAB = \angle DPB = 90^{\circ}$. Since $P$ is the midpoint of $CP$ (by construction) and $M$ is the mid-point of $CA$ (given), it follows that $PM$ is parallel to $DA$ and $DA = 2PM = BP$. Thus $DBPA$ is an isosceles trapezium and $DB$ is parallel to $PA$.

We hence get

$$\angle DPA = \angle BAP = \angle BCP = \angle NPC$$

the last equality follows from the fact that $\angle BPC = 90^{\circ}$, and $N$ is the mid-point of $CB$ so that $NP = NC = NB$ for the right-angled triangle $BPC$. It follows that $A, P, N$ are collinear.

Solution:

We use coordinate geometry. Let us take $P = (0, 0)$, and the coordinate axes along $PC$ and $PB$; we take $C = (c, 0)$ and $B = (0, b)$. Let $A = (u, v)$. We see that $N = (c/2, b/2)$ and $M = ((u + c)/2, v/2)$. The condition $PB = 2PM$ translates to

$$(u + c)^2 + v^2 = b^2$$

We observe that the slope of $CP = 0$; that of $CB$ is $-b/c$; that of $PA$ is $v/u$; and that of $BA$ is $(v - b)/u$. Taking proper signs, we can convert $\angle PCB = \angle PAB$, via tan function, to the following relation:

$$u^2 + v^2 - vb = -cu$$

Thus we obtain

$$u(u + c) = v(b - v), \quad c(c + u) = b(b - v)$$

It follows that $v/u = b/c$. But then we get that the slope of $AP$ and $PN$ are the same. We conclude that $A, P, N$ are collinear.