Solution:
Since the quadrilateral $B M E L$ is cyclic, we have $\angle B E M = \angle B L M$. By construction, $|B K| = |B L| = |B M|$, and so (using directed angles)
$$\begin{aligned} \angle B L M & = 90^\circ - \frac{1}{2} \angle M B L = 90^\circ - \left(180^\circ - \frac{1}{2} \angle L B K - \frac{1}{2} \angle K B M\right) \\ & = \left(\frac{1}{2} \angle L B K + \frac{1}{2} \angle K B M\right) - 90^\circ = \left(180^\circ - \angle B\right) - 90^\circ = 90^\circ - B. \end{aligned}$$
We see also that $\angle B E M = \angle B A H$, and so the point $N$ of intersection of $E M$ and $A H$ lies on $\Gamma$.
Let $X$ be the point of intersection of $K H$ and $B C$, and let $N'$ be the point of intersection of $M X$ and $A H$. Since $B C$ bisects the segment $K M$ by construction, the triangle $K X M$ is isosceles; as $A H \parallel M K$, $H X N'$ is isosceles. Since $A H \perp B C$, $N'$ is the reflection of $H$ in the line $B C$. It is well known that this reflection lies on $\Gamma$, and so $N' = N$. Thus $E, M, N$ and $M, X, N'$ all lie on the same line $M N$; that is, $E M$ passes through $X$.

Solution:
We work with directed angles. Let $H K$ meet $B C$ at $X$. Let $M X$ meet $A H$ at $H_{A}$ on $\Gamma$ (where $H_{A}$ is the reflection of $H$ in $B C$). Define $E'$ to be where $H_{A} M$ meets $\Gamma$ (again). Our task is to show that $\angle M E' B = \angle M L B$.
Observe that
$$\begin{aligned} \angle M E' B & = \angle H_{A} A B & \text{ (angles in same segment) } \\ & = B^{c} & \end{aligned}$$
Now
$$\begin{aligned} \angle M L B & = \angle H L B \\ & = \angle B K H_{C} \\ & = \angle B C H_{C} \\ & = B^{c} . \end{aligned}$$
(Simson line, doubled)
(reflecting in the line $A B$)
(angles in the same segment)