Let N be a point on the line $AK$ so that $MN \parallel AL$. Since $\frac{CM}{DL} = \frac{NM}{AL} = \frac{KM}{KL}$ and $\angle CMN = \angle DLA$, it follows that $K$ is a center of homothety which sends $\triangle CMN$ into similar $\triangle DLA$. On the other hand, $\triangle DLA$ is similar to $\triangle CBA$ because $\angle DAL = \angle BAC$ and $\angle ADL = \angle ACB$. Consequently, $\angle (BN, BC) = \angle (MN, MC)$, and thus points $N$, $B$, $M$, $C$ are cyclic. Therefore, $\angle CBM = \angle CNM = \angle CAB$ from which it follows that $BM$ touches $\omega$.

For this solution we need the following theorem.

Pascal's theorem. Points $A$, $B$, $C$, $D$, $E$, $F$ (not necessarily in this order) lie on some circle. Then the intersections of the lines $AB$ and $DE$, $BC$ and $EF$, $CD$ and $FA$ lie on a straight line.

Back to the problem. Let $E$ be the second intersection of the line $AL$ and $\omega$. Define $l_b$ as a tangent line to $\omega$ at $B$. Let's apply Pascal's theorem on points $B$, $B_1$, $A$, $E$, $C$, $D$ (here $B_1$ coincides with $B$) and pairs of lines ($BB_1$, $EC$), ($B_1A$, $CD$), ($AE$, $DB$). These pairs of lines coincide with pairs of lines ($l_b$, $EC$), ($BA$, $CD$), ($AE$, $DB$). Then from Pascal's theorem, the straight line connecting $K = BA \cap CD$ and $L = AE \cap DB$ will also contain $l_b \cap EC$. Since $M = EC \cap KL$, it follows that $BM$ touches $\omega$, as desired.