• Realizing $BCEF$ is cyclic.
• Relating $\angle EMK$ and $\angle EBK$ using $\angle FME$

$\angle BEC = \angle CFB = 90$

So, $BCEF$ is cyclic. Let this circle be $\Gamma$

$BC$ is the diameter of $\Gamma$ because $BEC = 90$

So, $M$ is the center of $\Gamma$

As $N$ is the midpoint of arc $EF$, $MN$ is perpendicular bisector of $EF$.

Now, $\angle EMK = \angle EMN = \frac{1}{2}\angle FME = \angle FBE = \angle KBE$

So, we have $\angle EMK = \angle EBK \implies EBKM$ cyclic.

(Fun fact: $A$ was not used in this solution.)