I will prove a more generalized problem In fact condition $PQ$ parallel to $BC$ is not needed $\angle CKS=\angle CKL=\angle CAL=\angle ACB=\angle AKB$ Hence $KS$ and $KA$ are isogonal in $\angle BKC$ Since $P$ is intersection of $CS$ and $BA$ and $Q$ is intersection of $BS$ and $CA$, by the isogonal line lemma $KP$ and $KQ$ are isogonal in $\angle BKC$ which is what we needed to prove

Attachments:

After proving that $KS$ and $KA$ are isogonal in $\angle BKC$ (*), one might finish with DDIT for the degenerate quadrilateral $BQCP$ as well: note that there is an involution at $K$ interchanging $(KA, KS), (KB, KC), (KP, KQ)$ and because of (*), this involution is a symmetry wrt the angle bisector of $\angle BKC$, so we are done.

Does anyone have another solution without using isogonality ?

I will use the condition $PQ \parallel BC$ $AS$ intersects $PQ,BC$ at $N,M$; $LS$ intersects $PQ,BC$ at $I,G$. Easy to see $M$ is the midpoint of $BC$. Also, $AM,CP,BQ$ are concurrent, therefore $(AS,NM)=-1$ Since $AL \parallel NI \parallel GM$, we have $(LS,IG)=(AS,NM)=-1$ $\Rightarrow (ML,MS,MI,MG)=-1$ Note that $MA=ML$, so $MG$ is the external bisector of $\angle{AML}$ Therefore, $MI$ is the internal bisector of $\angle{AML}$, or $MI$ is the perpendicular bisector of $BC$. $\Rightarrow \angle{BIM}=\angle{CIM} \Rightarrow \angle{BIP}=\angle{CIQ}$ Moreover, $\angle{BKI}=180^o-\angle{BAL}=180^o-\angle{BPQ}=\angle{BPI}$, hence $BPIK$ is cyclic. Analogously, $CQIK$ is cyclic, therefore $\angle{BKP}=\angle{BIP}=\angle{CIQ}=\angle{CKQ}$

https://www.matematickitalent.mk/uploads/books/Mam-5P2Z6Uivs9MldDbD4w.pdf The official solutions use much simpler methods (especially Solutions 2 and 3).