The angles $LCA$ and $LOA$ are equal as inscribed angles upon the same arc. The angle $LOA$ is equal to the sum of the angles $OAB$ and $OBA$, as an external angle to the triangle $ABO$, therefore:
$$\begin{aligned} \angle LCO &= \angle LCA + \angle OCA = \angle LOA + \angle OCA = \angle OAB + \angle OBA + \angle OCA = \\ &= \frac{1}{2}(\angle CAB + \angle ABC + \angle ACB) = 90^\circ \end{aligned}$$

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Analogously $\angle KCO = 90^\circ$, so the point $C$ lies on the line $KL$, and point $C$ is the middle point of $\overline{ON}$. The line $PC$ is parallel to $MN$ as a middle line of triangle $MON$.

The quadrilateral $LOKM$ is a parallelogram, because its diagonals bisect each other at point $P$. That implies that, the angles $MLK$ and $OKL$ are equal.

On the other hand, the triangle $OKN$ is isosceles with base $ON$ ($KC$ is its height and median), so $KC$ is a bisector of $OKN$, i.e. the angles $OKC$ and $NKC$ are equal. It follows that $\angle MLK = \angle NKL$, therefore the quadrilateral $KLMN$ is an isosceles trapezoid, hence inscribed. (If point $P$ lies on the other side of point $C$, we consider the angles $MKL$ and $NLK$).