Let $W'$ be the circumcenter of the triangle $ACP$, and let $L'$ be the intersection point of the line $FW'$ and the side $AC$. Note that the points $A, F, C$ and the foot $T$ of the perpendicular from $A$ to $PC$ are on the circle with a diameter $AC$, and thus $\angle PAT - \angle PCF$. Since $AC = BC$ and $F$ is the midpoint of $AB$, $\angle HAF = \angle HBF$, and so $\angle PBJ = \angle HCJ$, which along with $\angle PJB = \angle HJC$ induce that the triangles $PBJ$ and $HCJ$ are similar to each other. Therefore, for the feet of the perpendiculars from $J$ to $FP$ and $FC$, $\frac{XJ}{JY} = \frac{BP}{CH}$. Let $S$ be the foot of the perpendicular from $W'$ to $AP$. Then $HC = 2W'S$ and
$$BP = BF - PF = AF - (PF - FS) = AF - (AS - FS) = 2FS.$$
It follows that $\frac{XJ}{XF} = \frac{FS}{W'S}$. Hence $\triangle JXF$ is similar to $\triangle FSW'$, and thus $\angle JFX = \angle FW'S$. Since $W'S$ is parallel to $CF$, $\angle FW'S = \angle CFW'$, Thereby $\angle JFX = \angle CFW'$. Now, since $\angle PFC = 90^\circ$, $\angle JFL' = \angle JFW' = 90^\circ$, which along with $\angle MEL' = 90^\circ$ implies that the points $E$ and $F$ lie on the circle with a diameter $ML'$.

Suppose $AW = PW$, in which case $W$ is the circumcenter of $\triangle APC$, $AW = PW = CW$. From the fact we proved above, $E$ and $F$ lie on the circle with a diameter $ML$. Since the circumcenter $O$ of $\triangle ABC$ is the orthocenter of $\triangle AKC$, $AO$ is orthogonal to $KC$. But $BL$ is parallel to $KC$, and so $AO$ is orthogonal to $BL$. Thus $\angle ABL + \angle BAO = 90^\circ$. Note that $A, F, O, E$ are concyclic, and thus $\angle BAO = \angle MEF$, from which we can conclude $\angle FBL + \angle FEL = \angle ABL + \angle MEF + 90^\circ = 180^\circ$. This implies that $F, B, M, L, E$ are concyclic.

Conversely, suppose $B, E, F, M$ are concyclic. Since $E$ and $F$ lie on the circle with a diameter $ML'$, $\angle MBL' = 90^\circ$. It can be easily seen that $\angle MBK = \angle MEF = \angle BAO$, and so $BM$ is parallel to $AO$. It follows that $AO$ and $BL'$ are orthogonal to each other, thereby $BL'$ and $KC$ being parallel to each other. Therefore, $BL' = BL$, which implies $L' = L$ and $W' = W$. $\square$