Suppose that $AC \le BC$. Let $D$ be a point on the side $AB$ such that $DM \parallel BC$. Then
$$\frac{DB}{DA} = \frac{CM}{AM} = \frac{BL}{CL},$$
i. e. $DL \parallel AC$. On the tangent line to $\omega$ at point $C$ let's choose a point $T$, such that $KT \parallel BC$. Then $\angle TKA = \angle CBA = \angle TCA$. Therefore, $AKCT$ is cyclic. Let segments $KT$ and $AC$ intersect at point $S_1$. Then
$$\frac{S_1P}{S_1C} = \frac{KP}{KL} = \frac{KA}{KD} = \frac{S_1A}{S_1M} \implies S_1P \cdot S_1M = S_1C \cdot S_1A = S_1K \cdot S_1T.$$
Thus, $TMKP$ is inscribed into the circumcircle of triangle $KMP$. It is known that the common chords of three pairs of circles, centers of which are not collinear, are concurrent. It means that the common chord of $\omega$ and the circumcircle of $KMP$ passes through point $S_1$. Therefore, point $S_1$ coincides with $S$, and $S_1K \parallel BC$ follows from the definition of $T$.

Solution 2:

Let's introduce some notation:
$$b = AC, \ m = AM, \ s = AS, \ p = AP.$$
Since $S$ lies on the radical axis of the circumcircles of $ABC$ and $PKM$, then
$$SM \cdot SP = SA \cdot SC \implies (m-s)(s+p) = s(b-s) \implies s = \frac{pm}{b+p-m}.$$
So,
$$\frac{AS}{SC} = \frac{s}{b-s} = \frac{pm}{(b-m)(b+p)}.$$
According to Menelaus' theorem for triangle $ABC$ and transversal line $PKL$:
$$\frac{AK}{KB} \cdot \frac{BL}{LC} \cdot \frac{CP}{PA} = 1 \implies \frac{AK}{KB} = \frac{CL}{LB} \cdot \frac{AP}{PC} = \frac{AM}{MC} \cdot \frac{AP}{PC} = \frac{m}{b-m} \cdot \frac{p}{b+p} = \frac{AS}{SC} \implies SK \parallel BC,$$
Q. E. D.