Let $BC\cap AD=O$ then given condition implies $O,K,L$ are collinear. Given conditions also implies $BC$ is common tangent to circles $ABP,CQD$. Note that a homothety with center $O$ sends circles $ABP$ to circle $DQC$. Let $KL\cap \odot ABP =\{P,H\}$ and $KL \cap \odot \{G,Q\}$. Then property of homothety implies: $\frac{OH}{OQ}=\frac{OP}{OG}$...(i) . Power of point implies $(OB^2)(OC^2)=(OH.OP)(OQ.OG)=OP^2.OQ^2$ (using (i))
That means $OB.OC=OP.OQ$ Power of point theorem implies $P,Q,B,C$ are concyclic.
That means $OB.OC=OP.OQ$ Power of point theorem implies $P,Q,B,C$ are concyclic.