1975 AMC 12 Problem 11

Join a chord midpoint $M$ to the circle’s center $O$

Since $OM$ is perpendicular to the chord through $P,$ $\angle OMP=90^\circ$

If $M$ is a chord midpoint and $O$ is the center, then $OM$ is perpendicular to the chord, so $\angle OMP=90^\circ$ unless $M=O.$ Thus $M$ lies on the circle with diameter $OP.$ Conversely, each nonendpoint $M$ on that circle gives $PM\perp OM,$ so the line $PM$ cuts a chord of $K$ whose midpoint is $M.$ The endpoint $M=P$ gives the chord through $P$ perpendicular to $OP,$ while $M=O$ gives the diameter through $O$ and $P.$

Therefore, the correct answer is E.