Let the point $C$ be the midpoint of the line segment $AB$. We have to prove $MC \parallel KP$.
Let us introduce angle $\alpha = \angle BKA$. Notice that
$$\begin{aligned} \angle BLA &= 180^\circ - \angle BMA = 180^\circ - \frac{1}{2}\angle BOA = \\ &= 180^\circ - \frac{1}{2}(180^\circ - \angle BKA) = 90^\circ + \frac{1}{2}\alpha \end{aligned}$$
Also, notice that the point $O$ is midpoint of the arc $\widearc{AB}$. Thus the line $KO$ is bisector of the angle $\angle BKA$. From the two claims above, we deduce that $L$ is incenter of the triangle $ABK$.
Moreover, notice that $ML$ is diameter of the circle $k_2$, thus $\angle ABM = 90^\circ$. Since $BL$ is angle bisector of the angle $\angle ABK$ we deduce that $BM$ is exterior angle bisector of the same angle.
Thus, since $M$ lies on angle bisector $KM$ and exterior angle bisector $BM$, $M$ is the center of the excircle for the triangle $ABK$.

Thus, we have to prove that the line passing through the incenter $L$ of the triangle $ABK$ and point of the tangency of incircle of the same triangle is parallel to the line passing through the center of the excircle $M$ and the midpoint $C$ of the line segment $AB$. This is a well known lemma, which completes the proof.