let point M' is on YC and M'K is perpendicular to BK easy to see that BKM'Y is parallelogram so KM = BY = BX and let XF is perpendicular from X on YC easy to see that if XF intersect BC at point H then H is orthocenter of XYC and XH = KM' and XH || KM' so XK || M'H so HM' is perpendicular to BX and goes throw point H' were H' is intersection of XY and BC so M' = M

It's easy to see that $BX=BY,PM=YM,PK=XK$ and that $BX\perp KC, BY\perp MY$. We get $PM^2+BK^2=PM^2+BX^2+XK^2=YM^2+BY^2+PK^2=BM^2+PK^2$, thus $PB\perp MK$. This problem looks similar to http://www.artofproblemsolving.com/Forum/viewtopic.php?t=34205

This solution runs better when $X$ is closer to $C$ than $B$, but it is of no matter. Let $\Gamma$ be the circle and denote $L=BP\cap\Gamma$ and $R=BC\cap PM$. Since $BC$ is the diameter of $\Gamma$, we have $\angle CYB=90^{\circ}$ and therefore $\angle PBY=90^{\circ}$ since $BP||CY$ which implies $YL$ is a diameter of $\Gamma$. Then $\angle LXY=90^{\circ}$ and this means $XL||BC$. So $XCBL$ is a cyclic trapezium and therefore an isosceles trapezium and thus $\triangle KBC$ is isosceles with apex $K$. This implies $KL=KX$ and $KB=KC$. Easy angle chasing shows $\angle CMR=\angle BPM=\angle LKX$. Also $\angle MRC=\angle KCB=\angle BCY=\angle MCR$ so $\triangle MCR$ is $M$-isosceles. Also, clearly we have $KL=KX$ and it follows $\triangle KLX$ and $\triangle MCR$ are similar. To prove they are congruent, we can note that $MC=PK$ because of the parallelogram $PKCM$, but $\angle BPY=\angle PYC=\angle YXC=\angle PXK$ so $PK=KX=KL$. Hence $KL=MC$ and since $LK||MC$ this forces $KM||LC\perp BP$.

Johan Gunardi wrote: It's easy to see that $BX=BY,PM=YM,PK=XK$ and that $BX\perp KC, BY\perp MY$. From here you can also do $MY=PM=KC=KB,$ $MY||KB,$ so $MYBK$ is parallelogram, but $MY\perp BY,$ so it is a rectangle. Hence, $KM\perp BP$

Lemma: In triangle $ABC$, $D$ is an arbitrary point on segment $BC$. $E$ and $F$ are the intersection points of the perpendicular bisectors of $BD$ and $CD$ with $AB$ and $AC$ respectively. Let $O$ to be the circumcenter of triangle $ABC$. Then, $AEOF$ is concyclic. Let $O$ to be the circumcenter of triangle $CXY$.Using this lemma in the main problem, we get that $CKOM$ is concyclic. so $KMC=KOC$. $CY||PB$ So $KCB=KBC=BCY$. $O$ is the midpoint of $BC$, so $KOC=KMC=90$.And we are done.

$PM=YM$ [because $PM||XC$] so,$KC=YM$ $\angle KBC=\angle BCY=\angle KCB$ $\therefore KB=KC$ $KB=KC=MY$ $\therefore MYBK$ is a parallelgram. $\therefore \angle MKB=\angle MYB =90$ so, $MK$ is perpendicular to $PB$ .

We use barycentric coordinates on $\triangle ABC$. The circumcircle has equation $a^2yz+b^2zx+b^2xy=0$. Then, point $T$ is one of two points $(t:u:u)$ satisfying the circumcircle equation, so $a^2u^2+2b^2ut=0$ implying $T=(-a^2:2b^2:2b^2)$. Let $P=(0:y:z)$ with $y+z=1$ so $X=(z:y:0)$ and $Y=(y:0:z)$. Additionally scale so $4b^2-a^2=1$. Then the displacement vectors are $(-a^2:2b^2-y:2b^2-z)$ and $(z-y:y:-z)$. EFFT translates the desired to \[0=a^2(-2b^2z+yz+2b^2y-yz)+b^2(a^2z+(2b^2-z)(z-y))+b^2(-a^2y+(2b^2-y)(z-y)).\]Dividing through by $y-z$ (if it is $0$ we are done) reduces the desired to \[0=2a^2b^2-b^2a^2-b^2(4b^2-1)=b^2(a^2-4b^2+1),\]which is obvious.

∠BYC = 90 and CY || BK ---> ∠KBY = 90 ∠BCK + ∠XBC = 90 and ∠CBK + ∠YBC = 90 ---> ∠BCK = ∠CBK MY = MP = CK = BK ---> YBKM is rectangle ---> ∠MKB = 90