Iranian Mathematical Olympiad

Let $Q$ be the intersection point of $PK$, $AB$ and let $L'$ be the incenter of triangle $QSP$. We'll prove that $L \equiv L'$.

We know that $\widehat{QSP} = \widehat{OKP} = 90^\circ$ therefore $QKSO$ is a cyclic quadrilateral. And since $PS = PK$, it's also concluded that $PQ = PO$. Now we have
$$\widehat{QLS} = 90^\circ + \frac{1}{2}\widehat{QPS} = 90^\circ + \frac{1}{2}\widehat{QPO} = 180^\circ - \widehat{QOP} = 180^\circ - \widehat{QOS}.$$
So $QL'SO$ is a cyclic quadrilateral and points $Q, K, L', S, O$ lie on circle with diameter $OQ$. Note that
$$\begin{aligned} \widehat{QL'K} &= \widehat{QSK} = 90^\circ - \widehat{KSP} = \frac{1}{2}\widehat{KPS} = \widehat{KPL'} \\ \implies QK \cdot QP &= (QL')^2 \end{aligned}$$
The LHS of the latest equation is the power of point $Q$ with respect to $\omega_2$. Since $Q$ lies on the radical axis of circles $\omega_1, \omega_2$, we have
$$(QL')^2 = QK \cdot QP = QO^2 - QA^2 = (QL')^2 + (OL')^2 - OA^2$$
$$\implies OL' = OA.$$
So $L'$ lies on $\omega_1$ which means $L$ and $L'$ are the same points. Now since $L$ lies on the angle bisector of $\widehat{KPS}$, we finally conclude that $SL = KL$.