From $EC^2-EP^2=KC^2-KP^2$ and $FP^2-FB^2=KP^2-KB^2,$ we get $KB^2-KC^2+EC^2-EP^2+FP^2-FB^2=0.$ Since $AP \perp EF,$ then $FP^2-EP^2=FA^2-EA^2=FB^2-EC^2$ $\Longrightarrow$ $KB^2-KC^2+FB^2-EC^2+EC^2-FB^2=0$ $\Longrightarrow$ $KB=KC.$

Let $D$ be midpoint of $BC.$ Parallels to $BP, CP$ through $F, E$ meet at $J \in AH,$ on account of central similaity $\triangle AFE \sim \triangle ABC.$ Reflections of $EF$ in $EJ, FJ$ meet at $Q.$ Then $J, K$ are incenter and Q-excenter of $\triangle QFE.$ Incircle $(J)$ touches $EF$ at $S \equiv AH \cap EF$ and excircle $(K)$ at $T,$ such that $FT = ES$ $\Longrightarrow$ $K$ is on D-altitude of $\triangle DEF,$ identical with perpendicular bisector of $BC.$

Conjure points $B_1$ and $C_1$ on lines $PB$ and $PC$ such that $FB=FB_1$ and $EC=EC_1$. Note that $B_1$ and $C_1$ lie on circles with diameters $AB$ and $AC$ respectively. Since $P=BB_1 \cap CC_1$ lies on the radical axis of $(AB), (AC)$, we see that $B, C_1, B_1, C$ lie on a circle. Thus $KB=KB_1$ and $KC=KC_1$ implies that $K$ is the centre of the circle through $BC_1B_1C$, hence $KB=KC$.

Let $ D $ be the midpoint of $ BC $ and let $ K^* $ be the complement of the orthocenter of $ \triangle BPC $ WRT $ \triangle ABC $. From $ K^*E $ $ \perp $ $ CP $ and $ K^*F $ $ \perp $ $ BP $ we get $ K^* $ $ \equiv $ $ K $, so $ DK $ is perpendicular to $ BC $ $ \Longrightarrow $ $ KB $ $ = $ $ KC $.

Lets use dot product. Set $O$ the circumcentre as origin .From $KF\perp BP, EK\perp CP $ we get $$(\dfrac{a+c}{2}-k) . (c-p)=0$$$$(\dfrac{a+b}{2}-k) . (b-p)=0$$Subtract second from first and get $$(\dfrac{c-b}{2}) . (a-p) - k.(c-b)=0$$Now as $AP\perp CB$ we have $(c-b).(a-p)=0$ so that we must have $k.(c-b)=0$ that is $OK\perp BC$ and we are done.

Let $D$ be the midpoint of $\overline{BC}.$ Since the sidelines of the medial triangle $\triangle DEF$ are parallel to those of $\triangle ABC$, the perpendiculars from $B, C, P$ to $DF, DE, EF$ are concurrent at the orthocenter of $\triangle ABC.$ Therefore, $\triangle BCP$ and $\triangle ABC$ are orthologic. Hence, $K$ is the orthology center of $\triangle DEF$ w.r.t. $\triangle BCP$, so $KD \perp BC.$ Thus, $K$ lies on the perpendicular bisector of $\overline{BC}$, i.e. $KB = KC.$

Another solution on similar lines of dukejukem.. We see that the condition ensures that $P$ is orthology centre of $ABC$ wrt $KEF$ (perps from $A$ to $EF$ etc concur there). So the other orthology centre of $KEF$ wrt $ABC$ must be $O$ the circumcentre(bcoz perps from $E$ to $AC$ and from $F$ to $AB$ concur there). Hence $OK\perp BC$ and done.