Belarusian Mathematical Olympiad

Answer: $30^\circ$.

Since $L$ lies on the bisector of $\angle KCB$, $L$ is equidistant from the lines $CP$ and $CB$. Similarly, since $L$ lies on the bisector of $\angle KMB$, $L$ is equidistant from the rays $MK$ and $MB$. Therefore, $L$ is an equidistant point for the rays $KM$ and $KP$, so $L$ lies on the bisector of $\angle PKM$. Thus, $\angle PKL = \angle BKM = \angle MKC$. Therefore, each of these angles is equal to $60^\circ$. Let $\angle ACB = 3x$. Since $\angle LKC = 120^\circ$, we have $\angle KAC = \angle LKC - \angle ACK = 120^\circ - x$. Therefore
$$\angle ABC = 180^\circ - \angle BAC - \angle BCA = 180^\circ - (120^\circ - x) - 3x = 60^\circ - 2x.$$
So,
$$\begin{aligned} \angle LMB &= 0.5\angle KMB = 0.5(180^\circ - \angle LKM - \angle ABC) = \\ &= 0.5(180^\circ - 60^\circ - (60^\circ - 2x)) = 30^\circ + x. \end{aligned}$$
Since $\angle LMB = \angle LCM + \angle MLC$, we obtain $\angle MLC = 30^\circ + x - x = 30^\circ$.