Because PE is parallel to $BC$ and PD is parallel to $AC$, the two right angled triangles $APE$ and $PBD$ are similar. This implies that there exists a positive number $s$ (the similarity factor) such that
$$|AP| = s|BP|, \quad |AE| = s|DP| \quad \text{and} \quad |EP| = s|BD|.$$

Using these equations together with $|DP| = |CE|$ and $|EP| = |CD|$, which are true since $CEPD$ is a rectangle, the Theorem of Pythagoras for triangle $PBD$ implies the desired identity:
$$\begin{aligned} s|BP|^2 &= s|BD|^2 + s|DP|^2 \\ |BP| \cdot |AP| &= |BD| \cdot |CD| + |AE| \cdot |CE|. \end{aligned}$$