Using the concyclic points, we have
$$\angle CAQ = \angle CBQ = \angle DBP = \angle DEP$$
and
$$\angle QCA = \angle QBA = \angle PBE = \angle PDE.$$
This implies $\triangle QAC \sim \triangle PED$. Let
$$\frac{QA}{PE} = \frac{QC}{PD} = \frac{AC}{ED} = k.$$

Applying Ptolemy's theorem to the cyclic quadrilateral $EBDP$, we obtain
$$BD \times PE + BE \times PD = BP \times DE.$$
It follows that
$$\begin{align*} AQ + CQ &= kPE + kPD \\ &= \frac{k}{AC}(BD \times PE + BE \times PD) \\ &= \frac{k}{AC}(BP \times DE) \\ &= BP. \end{align*}$$