We prove the converse first. Let $O$ be the midpoint of $BC$, which is the center of the circle. Then $PO,QO$ are both angle bisectors of the angles $BPQ,CQP$. Since $\angle BPQ+\angle PQC+\angle QCB+\angle CBP=360$, we have $\angle BPO+\angle OPC+\angle QCO=180$, so $\angle QOC=\angle OPB$. Similarly, $\angle POB=\angle OQC$, thus $\triangle OQC\sim\triangle POB$, so $\frac{BP}{BO}=\frac{CO}{CQ}\implies BP\cdot CQ=\left(\frac{BC}{2}\right)^2$. Now suppose $BP\cdot CQ=\left(\frac{BC}{2}\right)^2$, then we let $Q'$ on $AC$ such that $PQ$ is tangent to the circle. Then by above, $BP\cdot CQ'=\left(\frac{BC}{2}\right)^2$, so $Q\equiv Q'$.