Let $\overline{AD}$ be altitude and let $M \in (DEF)$ be the midpoint of $\overline{AH}$. Define $K' = HK \cap (DEF)$, where $K'$ is on the arc $EF$ not containing $D$. Claim: $MK' \perp (DHK')$ Proof: Define $P = MD \cap EF, Q = MK' \cap EF$, and $R$ be the intersection of $MK'$ with the line through $H$ parallel to $EF$. Well-known (shooting lemma?) that $MP \cdot MD = MK' \cdot MQ$, so $PDQK'$ is cyclic. As $HR \parallel PQ$, Reim's gives $HDRK'$ cyclic. But $\angle RHK' = 90^\circ$ so the center of $(DHK')$ is on $MK'$, as desired. $\Box$ Let $AD$ hit $(ABC)$ again at $H'$. The (negative) inversion through $H$ with power $-HD \cdot HA$ sends $D, K', M \mapsto A, K, H'$ so in the above claim becomes $AK \perp (HH'K)$. But $DH = DH'$ so the center of $(HH'K)$ also lies on $BC$, i.e. $P$ is its center. Now $PH = PK$ follows.

Shorter: Take $D$ the reflection of $H$ in $\overline{BC}$ and let $A'$ be the antipode of $A$. Then $$\angle{HDK}=90^0+\angle{A'DK}=90^0+\angle{KAO}=90^0+\angle{HKP}$$ where the last equality holds because $HK\parallel AO.$ Also note that $PH=PD.$ Those two facts are enough to imply that $P$ is the center $(HDK)$ which yields $PK=PH.$

Let WLOG $AC\geq AB$. Let $AH\cap\odot (ABC)=(A,D)$. Let $O$ be circumcentre of $\odot (ABC)$. Let $OD\cap BC=J$. Now see that $JH\parallel AO$ and $AO\perp EF$. $\therefore JH\perp EF$. $\therefore J$ lies on $KH$. Now note that $\angle PKD=\angle AKD=90^{\circ}-B+C$. Also $\angle OJP=\angle OJC=\angle CDJ+\angle DCJ=C+90^{\circ}-B$. So $\angle OJP=\angle PKD\implies JDKP$ is cyclic. So $\angle PKH=\angle PKJ=\angle PDJ$ But now $\angle PDJ=\angle PHJ=\angle PHK$ as $D$ is reflection of $H$ in $PJ$. So $\angle PKH=\angle PHK$. So $PK=PH$ as desired.

Here is a solution that uses only angel chasing . Let $D$ denote the foot of the $A$-altitude onto $BC$ . Let $H' \equiv AH \cap \odot (ABC)$ . Let $Q \equiv HK \cap BC$ and $ J \equiv EF \cap HK$. Also $G \equiv AH \cap EF$ Denote by $O$ ,the circumcenter of $\odot (ABC)$ . Note that $HK \parallel AO$ . Let $A'$ be the $A$-antipode . WLOG $AC >AB$ . Note that $\angle{PKH}=\angle{AKH}=\angle{ACH}= 90+C-B$ Claim : $O,Q,H'$ are collinear. Proof : Let $Q' \overset {\text {def}}{\equiv} OH' \cap BC $ . Note that $$\angle{HQ'D}=\angle{DQ'H'}=\angle{OQ'C}=\angle{OH'A'}=\angle{OA'H'}=\angle{(AA',BC)} \implies HQ' \parallel AA' \implies \boxed {Q' \equiv Q} $$ Next note that $$ \angle{OQC}=\angle{(AA',BC)}=\angle{PKH'} \implies PQHK \text { is cyclic }$$ Now we are ready to finish . We seek to compute $\angle{PHK}$ Note that $\angle{PHK}=\angle{PQK}=\angle{HQD}=\angle{HGJ} = \angle{HGF} =90+C-B = \angle{PKH}$ So $PK=PH'=PH$ . Done . $\blacksquare$

Let $AH$ meet $(ABC)$ at $T$. Let line through $H$ perpendicular to $HK$ meet $AK$ at $S$ and $(ABC)$ at $X,Y$. It is pependicular to $AO$, so $AX=AY$. Then $\angle ATX = \angle AXH$. It means inversion $(A,AX)$ sends $H$ to $T$. It also sends $XY$ to $(ABC)$, so $S$ goes to $K$. To this inversion we have $(HTKS)$ - cyclic. With $KS$ being its diameter and $PH=PT$ it follows that $P$ - center of $(HTKS)$.

Let $AA'$ be diameter of $ABC$ and Let $AH$ meet $ABC$ at $H'$. Note that $\angle OAC = \angle 90 - B = \angle 90 - \angle AEF \implies AO \perp EF \perp HK$ so $HK || AA'$. Note that $\angle PKH = \angle KAA' = \angle 90 - \angle AA'K = \angle AH'K - \angle 90$. Assume $O'$ is center of $HH'K$ and $O'$ is not $P$ then $\angle HO'K = \angle 360 - 2\angle HH'K \implies O'KH = \angle HH'K - 90 = \angle PKH$ so $O'$ is $P$ so $PH = PK$.

Let $AH$ intersect circumcircle at $H'$ Take $P^*$ on $BC$ such that $AOH'P^*$ is cyclic. We will show that $P=P^*$ Let $AP^*$ intersect circumcircle at $K^*$ $\angle OK^*P = \angle OAP =\angle OH'P$ implies $OP$ is perpendicular bisector of $H'K^*$ So, $P^*$ is the circumcenter of $HH'K^* \Rightarrow P^*H=P^*H'=P^*K^*$ Let $OP^*$ intersect $AH$ at $Q$ $\angle K^*OQ=\angle H'OP^*=\angle H'AK^*$ means $AOK^*Q$ is cyclic $\angle P^*QK^*=\angle P^*QH$ Also $\angle P^*KQ=\angle P^*H'Q=\angle P^*HA$ means $HPK^*Q$ is cyclic and $\angle OAK^*=\angle OK^*A=\angle OQH = \angle P^*K^*H$ implies $HK^* \parallel AO \Rightarrow K^*=K$ and $P^*=P\Rightarrow PH=PH'=PK$

lemma: Let c1 , c2 be two circles with centers O1 , O2 and radii r1 ,r2 how r1=r2 .c1, c2 intersect at A,B . Let j be an arbitrary line passes O1 and intersect c1 at Y and c2 at X . k is a parallel line from O2 to j and k intersectc1 at S and c2 at Z so XYSZ is a isosceles trapeziod and if XS intersect YZ at P P is on line AB . proof:XO2=r2=r1=SO1 so O1SO2X is a isosceles trapeziod as a result O1O2=SX similarly O1O2=ZY so ZY=XS so XYSZ is a isosceles trapeziod and we have PX .PZ=PY .PS so p is on line AB . let the line drawn from A and parallel to HK intersect BHC circumcircle at D how D is in the side of BC that there is not A . by lemma we undrestand AHKD is a isosceles trapeziod and HK intersect BC at P so PK=PH

Attachments:

Let $AH \cap (ABC)=H_A$, let $AA'$ diameter in $(ABC)$, note that $PH=PH_A$ so now by angle chase we have: $$\angle HKP=\angle KAA'=\angle KH_AA'=\angle HH_AK-90 \implies P \; \text{circumcenter of} \; \triangle HH_AK$$Hence $PH=PK$ as desired, thus we are done .

Let $AH\cap BC=D,AH\cap (ABC)=R,HK\cap (ABC)=L,AL\cap BC=Q,T$ be the point on $(ABC)$ where $AT\parallel BC$. If $RT\cap BC=X$, then $\measuredangle XHR=\measuredangle HRX=\measuredangle ARO=\measuredangle OAR$ thus, $XH\parallel AO\perp EF$ which implies $X,H,K,L$ are collinear. So $(K,L),(B,C),(R,T)$ is an involution. Project this from $A$ onto $BC$ to get $(P,Q),(B,C),(D,BC_{\infty})$ is an involution. Hence $DP.DQ=DB.DC=DA.DR$ which yields $A,P,Q,R$ are concyclic. Since $H$ is the reflection of $R$ with respect to $PQ$, we see that $H$ is the orthocenter of $\triangle APQ$. If $U,V$ are the feet of the perpendiculars from $P$ to $AQ,HK$, then under the inversion centered at $H$ with radius $\sqrt{HA.HD}$, $V^*$ lies on $AU$ so $V$ swaps with $L$. Since nine point circle swaps with the circumcircle, we conclude that $V$ lies on the nine point circle which implies $V$ is the midpoint of $HK$ as desired.$\blacksquare$