Solution:

Let $M$ be the midpoint of $AF$ and let $O$ be the circumcentre of triangle $CPD$. Now construct $Q$ to be the point such that $CPDQ$ is a parallelogram, and let $R$ be the centre of this parallelogram (i.e. $R$ is the intersection of $PQ$ with $CD$, and also $R$ is the midpoint of $PQ$).

Note that $QD = CP = FP$ and $DP = PA$ and $\angle QDP = 180^{\circ} - \angle DPC = \angle FPA$. Therefore (by SAS) we have a pair of congruent triangles:
$$\triangle QDP \cong \triangle FPA.$$
Therefore $\angle MAP = \angle RPD$ and $AF = PQ$. Thus $AM = \frac{1}{2} AF = \frac{1}{2} PQ = PR$. Therefore (by SAS) we have another pair of congruent triangles:
$$\triangle MAP \cong \triangle RPD.$$
Therefore $\angle APM = \angle RDP$. Let $x = \angle APM$ so that $\angle CDP = \angle RDP = x$ also.

Since the angle subtended at the circumcentre is double the angle subtended at the circumference, we get $\angle COP = 2x$ (recall that $O$ is the circumcentre of $\triangle PCD$). Finally we get $\angle OPC = 90^{\circ} - x$ because $\triangle COP$ is isosceles. Putting this all together, we get
$$\angle OPM = \angle OPC + \angle CPA + \angle APM = (90^{\circ} - x) + 90^{\circ} + x = 180^{\circ}.$$
Therefore $\angle OPM$ is a straight line.