58th Ukrainian National Mathematical Olympiad

We extend the ray $AB$ further after $B$ to find the point $T$ such that $BT=PC$ (Fig. 1). Then, $\Delta TBC=\Delta CPA$ due to two equal length sides and the same angle between them. Hence, $TC=CA$. Analogously, $$AK=KB+PC=KB+BT=KT.$$ Therefore, in the isosceles triangle $ATC$ segment $KC$ is a median, which also makes it the altitude. From which follows that $KC\perp AB$, Q.E.D.

Fig. 1