jmc

The sum of the angle measures of an octagon is $180(8-2)=1080$ degrees, so each angle of a regular octagon measures $1080^{\circ }/8=135^{\circ }$. Therefore, $\angle BCD=135^{\circ }$, which means $\angle BCP=180^{\circ }-\angle BCD=45^{\circ }$. Similarly, $\angle PAB=45^{\circ }$. Since $\angle ABC=135^{\circ }$, the reflex angle at $B$ that is an interior angle of $ABCP$ has measure $360^{\circ }-135^{\circ }=225^{\circ }$. The interior angles of quadrilateral $ABCP$ must sum to $360^{\circ }$, so we have $$\begin{array}{rl}\angle P& =360^{\circ }-\angle PAB-(\text{reflex }\angle B)-\angle BCP\\ & =360^{\circ }-45^{\circ }-225^{\circ }-45^{\circ }=\boxed{45^{\circ }}.\end{array}$$

Note that we also could have solved this problem by noticing that $\overline{FC}\parallel \overline{HA}$, so $\angle APD=\angle FCD$. Since $\overline{CF}\parallel \overline{ED}$, we have $\angle FCD=180^{\circ }-\angle CDE=180^{\circ }-135^{\circ }=45^{\circ }$.