Since $ABCD$ is a parallelogram and $E, M$ are the midpoints of $BC, AD$ respectively, $EM \parallel AB,\; BE \parallel AD,\; ED \parallel BM$.
Given $BK \perp AB \Rightarrow BK \perp EM$.
Similarly, $EK \perp ED$ and $ED \parallel BM \Rightarrow EK \perp BM$.
Thus $BK, EK$ are altitudes of $\triangle BEM$, hence $K$ is its orthocenter.
Therefore $MK \perp BE$.
Since $BE \parallel AD \Rightarrow MK \perp AD \Rightarrow MK \perp MD$, so $\angle KMD = 90^\circ$.
Also $EK \perp ED \Rightarrow \angle KED = 90^\circ$.
Hence $\angle KED = \angle KMD = 90^\circ \Rightarrow K, E, D, M$ are concyclic.
