Take $X,Y$ reflections of $B$ about $P,Q$ respectively and $Z$ reflection of $C$ about $P$.Now see that $\triangle ZYC$ is equilateral triangle, so $ZY=ZC=ZD$, $Z$ is circumcenter of $\triangle CXY$, thus $\angle CXY=30^\circ$, but $PQ$ is midline of $\triangle CXY$, so $PQ\parallel XY$, that's $\angle CPQ=120^\circ$.