Solution:
We use standard notation for the angles of triangle $A B C$. Let $P$ be the midpoint of $C H$ and $O$ the centre of $\Gamma$. As
$$\alpha = \angle B A C = \angle B D C = \angle D K L$$
the quadrilateral $A C K L$ is cyclic. From the relation $C B = C D$ we get $\angle B C D = 180^\circ - 2 \alpha$, so
$$\angle A C K = \gamma + 2 \alpha - 180^\circ$$
where $\gamma = \angle A C B$. From the relation $C K = C A$ we get
$$\angle A L C = \angle A K C = 180^\circ - \alpha - \frac{\gamma}{2}$$
and thus from the triangle $A C L$ we obtain
$$\angle A C L = 180^\circ - \alpha - \angle A L C = \frac{\gamma}{2}$$
which means that $C L$ is the angle bisector of $\angle A C B$, thus $\angle A C L = \angle B C L$. Moreover, from the fact that $C H \perp A B$ and the isosceles triangle $B O C$ has $\angle B O C = 2 \alpha$, we get $\angle A C H = \angle B C O = 90^\circ - \alpha$. It follows that,
$$\angle N P H = 2 \angle N C H = \angle O C H$$

On the other hand, it is known that $2 C P = C H = 2 O M$ and $C P \parallel O M$, so $C P M O$ is a parallelogram and
$$\angle M P H = \angle O C H$$
Now from (3) and (4) we obtain that
$$\angle M P H = \angle N P H,$$
which means that the points $M, N, P$ are collinear.