Since $AM\perp BM \, , \, CL\perp DL$ and $AN\perp BN \, , \, CK \perp DK$, then it holds $$ \left\{\begin{array}{lll} BM\perp CM \, , \, BL\perp CL \Longrightarrow BMLC \quad \text{cyclic} \\ AN\perp DN \, , \, AK\perp DK \Longrightarrow ADKN \quad \text{cyclic} \\ \end{array} \right. $$Since $AD\parallel BC$, and quadrilaterals $BMLC$ and $ADKN$ are cyclic, then $$ \left\{\begin{array}{lll} \measuredangle KAD=\measuredangle MCB=\measuredangle MLN \\ \measuredangle KAD=\measuredangle KNL \end{array} \right. $$Hence $\measuredangle MLN=\measuredangle KNL$, and so $NK\parallel ML$.