First, let's try to characterize $P$. Let $Q$ be the isogonal conjugate of $P$ w.r.t. $\triangle ABC$. Note that $\angle QAC = \angle PAB = \angle ABB_m$ and $\angle QCA = \angle PCB = \angle B_mBC$. This means, $Q$ is the $B$-humpty point of $\triangle ABC$, i.e. $P$ is the $B$-dumpty point of $\triangle ABC$. Obviously, $Q$ lies on $BB_m$. Let $E'$ be the midpoint of $BQ$. We obtain \begin{align*} & \angle C_mE'B_m = \angle BC_mE' + \angle C_mBE' = \angle BAQ + \angle PAB = \angle PAC + \angle PAB = \angle BAC = \angle C_mA_mB_m \\ & \implies E' \in \odot(A_mB_mC_m). \end{align*}Let the $B$-symmedian of $\triangle ABC$ hit $\odot(ABC)$ at $D$. It's known that $P$ is the midpoint of $BD$, and $AQCD$ is a kite. Also, $E'P \parallel QD \implies E'P \perp AC \implies E' \equiv E$. So, $E \in \odot(A_mB_mC_m)$, as desired. $\square$

Nice. Let $T$ be on $ABC$ such that $BT$ is symmedian. Let $AP$ and $CP$ meet $ABC$ at $X$ and $Y$. Note that $\angle YAX = \angle BAP + \angle BCP = \angle B$ so $AY \parallel CX$. Also $\angle YAB = \angle PCB = \angle CBB_m = \angle ABT$ so $AY \parallel BT \parallel CX$ and since $AX,BT,CY$ are concurrent we have that $P$ lies on $BT$ and since $\angle PAY = \angle PYA = \angle B$ then $P$ is the midpoint of $BT$. Let $T'$ be reflection of $T$ w.r.t $AC$. It's well-known that $T'$ lies on $BB_m$. Now since $TT' \perp AC \perp PE$ and $P$ is midpoint of $BT$, we have that $E$ is the midpoint of $BT'$. Now $\angle C_mEA_m = \angle AT'B = 180 - \angle B = 180 - \angle C_mB_mA_m$ so $E$ lies on $A_mB_mC_m$.