Invert with centre $A$. Then $\angle AB'P'=\angle AC'Q'$ and $\angle AC'P'=\angle AB'Q'$, so $\angle P'B'Q'=\angle AB'P'-\angle AB'Q'=\angle AC'Q'-\angle AC'P'=\angle P'C'Q'$, which means that $B'C'Q'P'$ is cyclic. We also have that $K',P',Q',L'$ are collinear, so $\angle L'Q'C'=\angle P'B'C'$. Finally, since $\angle K'B'P'=180^\circ -\angle AB'P'=180^\circ -\angle AC'Q'=\angle Q'C'L'$, we get that $\angle C'L'K'=\angle C'L'Q'=\angle C'B'A$. Hence $B',C',L',K'$ are concyclic, so $B,C,L,K$ are concyclic, as desired.

My solution during the exam. Let the lines $AP$ and $AQ$ intersect $BC$ at $X$ and $Y$ respectively. Claim. $B,P,Q,C$ are concyclic. Proof. From the given we obtain $\angle{APB}+\angle{APC}=\angle{AQB}+\angle{AQC}\Rightarrow \angle{BPC}=\angle{BQC}.$ This proves the claim. Claim. $P,Q,Y,X$ are concyclic. Proof. Let $\angle{APB}=\angle{AQC}=a$ and $\angle{PBC}=b$. $\angle{PXY}=180-\angle{PXB}=\angle{BXP}+\angle{PBX}=180-a+b$ and $\angle{PQX}=\angle{PQC}-\angle{YQC}=a-b.$ If the two angles are summed we arrive at the conclusion. Claim. $K,P,X,B$ and $L,Q,Y,C$ are both concyclic. Proof. $\angle{PXY}=\angle{AQP}=\angle{BKP}\Rightarrow K,P,X,B$ are concyclic. ($A,K,P,Q$ concyclic) The other concyclicity can be proven similiarly. Claim. $B,C,L,K$ are concyclic. Proof. Invoking PoP from $A$ we obtain, $$AK\cdot AB=AP\cdot AX=AQ\cdot AY=AL\cdot AC.$$This finishes the proof. $\blacksquare$