Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

$\angle BEC = \angle CFB = 90$

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.)