Denote $\alpha = \langle CAB$, $\beta = \langle ABC$, $\gamma = \langle BCA$.

The centre of circle $k$ is the intersection of the bisectors of angles $\angle CBL$ and $\angle BCM$.
Therefore, it suffices to show that those bisectors intersect line $LM$ at points $P$ and $Q$ respectively.
Denote by $P'$ and $Q'$ the intersections of line $LM$ with bisectors of angles $\angle KBL$ and $\angle KCM$, respectively. We have to show that $P' = P$ and $Q' = Q$.
Triangles $KBP'$ and $LBP'$ are congruent because they have a common side $\overline{BP'}$, and $|KB| = |LB|$, $\angle KBP' = \angle LBP'$. It follows that $\angle P'KB = \angle BLP = \angle ALM$.
Triangle $ALM$ is isosceles, so $\angle AMP' = \angle AML = \angle ALM = \angle P'KB = 180^\circ - \angle P'KC$. From this it follows that quadrilateral $MCKP'$ is cyclic.
Since $|MC| = |KC|$, by the inscribed angle theorem we have $\angle MP'C = \angle KP'C$.
Since we also have $\angle KP'B = \angle LP'B$, it follows that
$$180^\circ = \angle MP'C + \angle KP'C + \angle KP'B + \angle LP'B = 2(\angle KP'C + \angle KP'B) = 2\angle CP'B,$$
therefore the angle $\angle CP'B$ is right. From this we can conclude that point $P'$ lies on the circle with diameter $\overline{BC}$.
Analogously, we can show that $Q'$ lies on the circle with diameter $\overline{BC}$.
Now we have $P', Q' \in \{P, Q\}$, since both $P'$ and $Q'$ are the intersections of line $LM$ and the circle with diameter $\overline{BC}$. This is possible only if $P = P'$ and $Q = Q'$; otherwise the centre of circle $k$ would be on the same side of line $LM$ as point $A$.
This completes the proof.