Let $M$ be the midpoint of $BC$ and $S$ be the reflection of $P$ over $M$. First we show that $S$ lies on $OH$.

We have
$$\begin{cases} PM = MS \\ PK = KA \end{cases} \Rightarrow MK \parallel AS, MK = \frac{AS}{2}.$$

$$\triangle MKO \sim \triangle AHS.$$
Because of the facts that $MK \parallel AS$ and $MO \parallel AH$, we conclude that $KO \parallel HS$ which means $S$ is a point on $OH$.

Now we show that $\overrightarrow{ESF} = 90^\circ$. Note that since $MB = MC$ and $MS = MP$, $BSCP$ is a parallelogram, therefore
$$\begin{cases} \overrightarrow{BSC} = \overrightarrow{BPC} = 180^\circ - \hat{A}, \\ \overrightarrow{SBC} = \overrightarrow{PCB} \Rightarrow \overrightarrow{SBE} = \hat{B} - \overrightarrow{PCB}, \\ \overrightarrow{SCB} = \overrightarrow{PBC} \Rightarrow \overrightarrow{SCF} = \hat{C} - \overrightarrow{PBC}. \end{cases}$$
So we have
$$\begin{cases} BS = PC = BE \Rightarrow \overrightarrow{ESB} = 90^\circ - \frac{SBE}{2} = 90^\circ - \frac{\hat{B}-\overrightarrow{PCB}}{2}, \\ CS = PB = CF \Rightarrow \overrightarrow{FSC} = 90^\circ - \frac{SCF}{2} = 90^\circ - \frac{\hat{C}-\overrightarrow{PBC}}{2}. \end{cases}$$
Therefore
$$\begin{align*} \overrightarrow{ESF} &= 360^\circ - \overrightarrow{ESB} - \overrightarrow{FSC} - \overrightarrow{BSC} \\ &= 360^\circ - \left(90^\circ - \frac{\overrightarrow{B-PCB}}{2}\right) - \left(90^\circ - \frac{\overrightarrow{C-PBC}}{2}\right) - \left(180^\circ - \hat{A}\right) \\ &= \frac{\overrightarrow{B}}{2} + \frac{\overrightarrow{C}}{2} + A - \frac{\overrightarrow{PCB}+\overrightarrow{PBC}}{2} \\ &= \frac{\overrightarrow{B}}{2} + \frac{\overrightarrow{C}}{2} + \frac{\overrightarrow{A}}{2} = 90^\circ. \end{align*}$$

Let $T$ be the circumcenter of triangle $ESF$, since $BT$ and $CT$ are the perpendicular bisectors of $ES$ and $FS$ respectively, and since $\overline{ESF} = 90^\circ$ we obtain $\widehat{BTS} = 90^\circ$.

Now we show that $MT \parallel AP$. Let $R$ be the intersection of $AK$ with $BC$. It suffices to show that $\overline{TMB} = \overline{ARB}$.
If $\widehat{PBC} = \alpha$ we have
$$\left. \begin{array}{l} \overline{ARB} = \alpha + \overline{APB} = \alpha + \hat{C}, \\ \widehat{BTC} = 90^\circ \\ MB = MC \end{array} \right\} \implies MB = MC = MT,$$
and so
$$\overline{TMB} = 2\overline{TCM} = 2\left(\overline{SCM} + \overline{TCS}\right) = 2\left(\alpha + \frac{\hat{C}-\alpha}{2}\right) = \alpha + \hat{C} = \overline{ARB}.$$
Therefore, $MT \parallel AP$. We have $\overline{SKP} = 90^\circ$. Also $MP = MS$, so $M$ is the circumcenter of triangle $SKP$ and thus it lies on the perpendicular bisector of $KS$. Also since $MT \parallel AP$, and since $AP \perp SK$ we conclude that $TM$ is the perpendicular bisector of $KS$ and so $TS = TK$. We also had $TS = TF = TE$. Therefore, points $E, S, K$ and $F$ lie on a circle with center $T$. So finally we obtain $\overline{ESF} = \overline{EKF} = 90^\circ$.