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$BC$ is a diameter of a circle and the points $X,Y$ are on the circle such that $XY \perp BC$.The points $P,M$ are on $XY,CY$(or their stretches),respectively,such that $CY \parallel PB$ and $CX \parallel PM$.Let $K$ be the meet point of the lines $XC,BP$.Prove that $PB \perp MK$.
(An easy problem.Just a little angle chasing)

are you sure about the statement?

tanmoy wrote: $CX \parallel PM$.

this line is confusing.

opsssss sorry my mistake

here is the solution:
$PM=YM$ [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 $PB$ .

Tahmid wrote:opsssss sorry my mistake

here is the solution:
$PM=YM$ [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 $PB$ .

Same as my solution.