This is equivalent to constructing $P$ such that there exist a point $D$ on $AB$ where $BDQP$ is a parallelogram and $PD\perp BC$. But I'm not quite sure how to do that.

Consider $R$ the symmetric of $Q$ in $P$ then we ll have $PQ.PR=PB.PC$ i.e $QRBC$ is cyclic more since $RQ \perp QC$ then $\angle RBC =90^\circ $ . let $A'$ the symmetric of $A$ in $B$ ,remark that it 's easy to show that $R$ is on $A'C$ thus $I$ the center of $(BCQR)$ is on the intersection of $A'C $ and the $BC$-bisector hence to construct $P$ draw the circle of center $I$ passing through $B$ ,it cuts $AC$ at a second point which is $Q$ then draw the perpendicular to $AC $ through which cuts $AB$ at the desired point RH HAS