WLOG $\angle B > \angle C$. It suffices to show that $d(K,BH) = \frac{1}{2} d(P,BH)$. Trivially, we have $AP=AC$ by simple angle chase, so $d(P,BH) = BP \cdot \sin \angle ABH = (b-c) \cos A = 2R (\sin B- \sin C) \cos A$. Now, doing some angle chasing and calculation, we find $d(K, BH) = HK \cdot \sin \frac{1}{2} A = AH \cdot \sin (90-C-\frac{1}{2}A) \cdot \sin \frac{1}{2}A = 2R \cos A \cdot \sin (90-C-\frac{1}{2}A) \cdot \sin \frac{1}{2}A$. Now, plugging stuff in, it suffices to show that $2 \sin (90-C-\frac{1}{2}A) \sin \frac{1}{2}A = \sin B - \sin C$, which is easy by the sum of trigonometric functions formula.

My solution : Let $BB_0, CC_0$ be the altitudes of $ABC$. $M$ be the midpoint of $BC$. $M_a$ be the midpoint of arc $BC$ not containing $A$. $N$ be the midpoint of $BP$. $T$ be the midpoint of $AH$. We will prove that $NK$ is parallel to $BH$. Step 0. $M_a$ is the circumcenter of $BIC$. This is well-known. Step 1. $MK = MM_a$. Firstly, $AB_0C_0H$ is cyclic with circumcenter $T$ and diameter $AH$. It is also clear that $K$ lies on circle with diameter $AH$, so $A,B_0,C_0,H,K$ are concyclic. Since $AK$ is the angle bisector of $\angle AB_0C_0$, then $K$ is the mid of arc $B_0C_0$ of circle $AB_0C_0$, so $KB_0 = KC_0$. But we have $MB_ 0 = MC_ 0$ and $TB_0 = TC_0$. So, $M,K,T$ are collinear. Because $AH$ and $MM_a$ are parallel, $\triangle{AHK} \sim \triangle{M_aMK}$ so $MK = MA_a$. Step 2. $MN \perp AM_a$ Clearly, $N,M$ are the projection of $M_a$ on $AB,BC$. Thus $MN$ will be the Simson line of $M_a$. So it passes the projection of $M_a$ the $AC$, say $Q$. But $NQ$ must be perpendicular to $AM_a$ since $M_a$ lies on angle bisector of $A$. So, step 2 is proved. Step 3. $NK$ paralel to $BH$. From step 1,2 we see that $NI = NM_a$. So \[\angle KPM_a = 2\angle MPM_a = 2\angle M_aBC = \angle BAC.\]So, $\angle APK = 90^{\circ} - \angle BAC = \angle ABH$. And step 3 is proved. Step 4. $B,H,Q$ collinear. Since $N$ is the mid of $BP$ and $NK$ parallel to $BH$ then the reflection of $P$ w.r.t $K$ lie on $BH$. And we are done.

Nice solution Meine, I will give mine With angle chasing, we can get $AI{\perp}PC$ meaning $\bigtriangleup{PAC}$ is isosceles. Suppose the intersection of $HB$ and $CP$ is $X$, the intersection of $AI$ and $CP$ is $D$ and the reflection of $H$ to $K$ is $H'$. $\angle{HKA}=\angle{PDA}=90^{\circ}$ so we have $HH'{\parallel}PC$ since $PD=DC$ and $HK=KH'$ we have $PH'=CH$. with some more angle chasing, we get $\angle{HXC}=\angle{HCX}$ resulting $HX=HC=PH'$ , since $XP{\parallel}HH'$ we have $XPH'H$ is a parallelogram. so $HH'=XP=2.HK$ so the intersection of $XH$ and $PK$, we call $T$, makes $\bigtriangleup{TPX}$$\simeq$$\bigtriangleup{TKH}$, since $XP=2.HK$, we have $K$ is the midpoint of $PT$, also meaning that $T=Q$

For obvious reasons, $\angle{ICB}=\angle{ICA}$, thus $AP=AC$, hence $AI\perp PC\implies KH\parallel BC$. Also, $PK=KC=KQ\implies \angle PCQ=90^{\circ}$, thus $KH\perp QC$. Let $R=KH\cap AC$. $KH$ bisects the angle $\angle CKQ$, since $KH$ is perpendicular bisector of $CQ$. Also, this implies that $\triangle QHC$ is isosceles. Hence, $R$ lies on $KH$, thus $QRC$ is isosceles. Now, $\angle{BAC}=2\angle{RCQ}=\angle{ARQ}\implies QR\parallel AB\perp CH$. Hence, $R$ is the orthocentre of $\triangle HQC$, thus $AC\perp HQ$ and because $BH\perp AC$, we conclude that $B$, $H$ and $Q$ are collinear.

Other write-up with asy It is obvious that $AP=AC$, this is due to trivial angle chase. Let $R$ be the reflection of $P$ over $A$. Thus, we have $AP=AR=AC$, hence $\measuredangle RCP=90^\circ$. Hence, $K$ is the circumcentre of $\triangle PCQ$. As $KH\perp AI\perp PC$, we get that $KH$ is the perpendicular bisector of $CQ$. Let $D$ be the intersection of $HQ$ and $AC$, we claim that $AKHD$ is cyclic, from which the desired collinearity follows. Indeed, $$\measuredangle DHK=\measuredangle KHC=\measuredangle PCH=\measuredangle CPA-90^\circ=\measuredangle ACP-90^\circ=\measuredangle DAK.$$We are done. [asy][asy] size(9cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; pair O,A,B,C,I,H,K,P,R,Q,D; O=(0,0);A=dir(60);B=dir(200);C=dir(340);I=incenter(A,B,C);H=orthocenter(A,B,C);path w=circle(A,abs(A-C));K=foot(H,A,I);P=intersectionpoints(w,B--2A-B)[1];R=intersectionpoints(w,B--2A-B)[0];Q=extension(P,K,R,C);D=extension(H,Q,A,C); draw(A--B--C--cycle,deep);draw(A--foot(A,P,C),deep);draw(P--C,deep);draw(P--Q,deep);draw(A--R--C,deep);draw(circumcircle(P,Q,C),med);draw(K--C,deep);draw(K--foot(K,C,Q),light);draw(circumcircle(A,K,H),light+dashed);draw(H--Q,med);draw(H--C,med); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$I$",I,dir(I)); dot("$H$",H,dir(H)); dot("$K$",K,dir(K)); dot("$P$",P,dir(P)); dot("$R$",R,dir(R)); dot("$Q$",Q,dir(Q)); dot("$D$",D,dir(D)); [/asy][/asy]