My solution: Let $ X=AC \cap BD $ . Let $ O $ be the center of $ \odot (ABCD) $ . Since $ \angle XOA'=\angle ACA'=\angle XCA'$ , so we get $ A', C, O, X $ are concyclic and $ PB \cdot PD=PX \cdot PO $ , hence from $ BO=OD \Longrightarrow (P,X;B,D)=-1 \Longrightarrow \tfrac{PB}{PD}=\tfrac{XB}{XD} $ . ... $ (1) $ Since $ XQ $ is the external bisector of $ \angle DXB' $ , so we get $ \tfrac{QB'}{QD}=\tfrac{XB'}{XD}=\tfrac{XB}{XD} $ . ... $ (2) $ From $ (1), (2) \Longrightarrow PQ \parallel BB' \Longrightarrow PQ\perp AC $ . Q.E.D

Let $E \equiv AC \cap BD.$ By obvious symmetry the tangents of $\odot(ABCD)$ at $A$ and $A'$ intersect on $BD$ $\Longrightarrow$ $ABA'D$ is harmonic $\Longrightarrow$ pencil $C(A,A',B,D)=-1$ is harmonic $\Longrightarrow$ $(E,P,B,D)=-1.$ But since $QE$ bisects $\angle BQB' \equiv \angle BQD$ $\Longrightarrow$ $QE,QP$ bisect $\angle BQD$ $\Longrightarrow$ $PQ \perp AC.$