Let: $(APD)=o_{1}$ $(BPD)=o_{2}$ $(APC)=o_{3}$ $(BQC)=o_{4}$ Let $M$ be the point of the intersection of line $DK$ with line $AB$. Because $M$ lies on a radical axis of $o_{1}$ and $o_{2}$ it's easy to conclude from the given relation that $M$ is a midpoint of $AB$ (since $|MP|\cdot|MA|=|MQ|\cdot|MB|)$. Analogically, $M$ lies on a radical axis of $o_{3}$ and $o_{4}$, and it's obvious that it lies also on the radical axis of $o_{1}$ and $o_{3}$ and on the radical axis of $o_{2}$ and $o_{4}$. So it's clear that $M$ is a radical center of these $4$ circles and in particular we have: $|MK|\cdot|MD|=|ML|\cdot|MC|$ which implies that $D, K, C, L$ is a cyclic quadrilateral (from the converse of the Power of the Point theorem).

more generally $AP=BQ$ isn't necessary.

We denote as $(O_{1},$ $(O_{2},$ $(O_{3},$ $(O_{4}),$ the circumcircles of the triangles $\bigtriangleup APD,$ $\bigtriangleup BQD,$ $\bigtriangleup APC,$ $\bigtriangleup BQC,$ respectively. So, we have $K\equiv (O_{1})\cap (O_{2}),$ $L\equiv (O_{3})\cap (O_{4})$ and let $E,$ $F$ be, the intersection points of $(O_{2}),$ $(O_{3})$ $($ In my drawing the point $P,$ is between $A,$ $Q$ and $F,$ between $E$ and side line $AB$ of $ABCD$ $).$ Let $M$ be, the intersection point of $AB,$ from the segment line $DK,$ as the radical axis of $(O_{1}),$ $(O_{2}).$ So we have that $(MP)\cdot (MA) = (MK)\cdot (MD) = (MQ)\cdot (MB)$ $,(1)$ From $(1)$ we conclude that the point $M,$ lies on the segment line $EF,$ as the radical axis of the circles $(O_{2}),$ $(O_{3}).$ Let $M'$ be, the intersection point of $AB,$ from the segment line $CL,$ as the radical axis of $(O_{3}),$ $(O_{4}).$ So we have that $(M'P)\cdot (M'A) = (M'L)\cdot (M'C) = (M'Q)\cdot (M'B)$ $,(2)$ From $(2)$ we conclude that the point $M',$ lies also, on the segment line $EF.$ From $(1),$ $(2)$ $\Longrightarrow$ $M'\equiv M$ and that the segment lines $DK,$ $EF,$ $CL,$ are concurrent at one point, here the point $M,$ lies on the side line $AB.$ So, we have that $(MK)\cdot (MD) = (ML)\cdot (MC)$ $,(3)$ From $(3),$ we conclude that the points $C,$ $D,$ $K,$ $L,$ are concyclic and the proof is completed. Kostas Vittas.

Umut Varolgunes wrote: more generally $AP=BQ$ isn't necessary. Yes! Doing inversion to $A$ the radius isn't necessary for this problem hence the problem becomes like this Image not found and we know that $D',E',F',C'$ is concyclic because $DC$ becomes a circle by inversion. Then we will show that $K'\in{B'L'}$ we know that $K'E'.K'D'=K'F'.K'C'$ because $E',D',C',F'$ are concyclic and $K'$ has equal power to circles and $BL$ is the radical axis of the circles so $K'\in{B'L'}$ line hence $K',B',L'$ is collinear so $A,B,K,L$ are concyclic. EDIT: In the orginal problem $P$ is $E$, $Q$ is $F$, $A$ is $D$, $B$ is $C$

Yet another PoP problem involves coincide. Let $AK\cap CD=P$ and $BL\cap CD=Q$. Observe that $P$ and $Q$ are radical axises of $(ADEK)-(AKFC)$ and $(BLED)-(BLFC)$, respectively. It directly implies that $PE\cdot PD=PF\cdot PC$ and $QE\cdot QD=QF\cdot QC$, we get the $P\equiv Q$ coincide (why?). Then by PoP $PE.PD=PK.PA=PF.PC=PL.PB$ which means $L\in (ABK)$ as desired.