Since $AB \parallel CD$, we have that $ABCD$ is isosceles trapezium. Let $O$ be the center of $k$ and $EM$ meets $AB$ at point $Q$. Then, from the right angled triangle $BEQ$, we have $BE^2 = BO \cdot BQ$. Since $BE = BK$, we get $BK^2 = BO \cdot BQ$ (1). Suppose that $KL$ meets $AB$ at $P$. Then, from the right angled triangle $BAK$, we have $BK^2 = BP \cdot BA$ (2).

From (1) and (2) we get $\dfrac{BP}{BQ} = \dfrac{BO}{BA} = \dfrac{1}{2}$, and therefore $P$ is the midpoint of $BQ$ (3). However, $DM \parallel AQ$ and $MQ \parallel AD$ (both are perpendicular to $DC$). Hence, $AQMD$ is parallelogram and thus $MQ = AD = BC$. We conclude that $QBCM$ is isosceles trapezium. It follows from (3) that $KL$ is the perpendicular bisector of $BQ$ and $CM$, that is, $M$ is symmetric to $C$ with respect to $KL$. Finally, we get that $M$ is the orthocenter of the triangle $DLK$ by using the well-known result that the reflection of the orthocenter of a triangle to every side belongs to the circumcircle of the triangle and vice versa.