Note that $M$ is the circumcenter of $(BCEF)$, so $MN$ bisect $\angle FME$. We have $$\angle KME=\frac{1}{2}\angle FME=\angle EBF.$$

It is well known that $ME$ and $MF$ are tangent to $(AFE)$. Hence, $ME = MF$. $\triangle MEF$ is isosceles, and $N$ is the midpoint of the nonequal side. So, $MN \perp EF$. (More specifically, $\triangle MEN \cong \triangle MNF$) Again, from three tangents lemma we get, $\angle MEF = \angle A \implies NME = \angle KME = 90^{\circ} - \angle A = \angle FBE$. So, $B, K, M, E$ is cyclic.