Take point $Q$ arbitrarily on the circle. Draw a diameter perpendicular to the fixed diameter, one of whose endpoints is $R$ (see the figure).

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We know that $O P=Q Q^{\prime}$. But $\angle R O Q = \angle O Q Q^{\prime}$, because they are alternate interior angles, and $O Q = O R$. Therefore, $\triangle O Q^{\prime} Q \cong \triangle O P R$, and thus $\angle R P O = 90^{\circ}$, which means that segment $O R$ is seen from point $P$ at a right angle. The geometric locus of points from which a segment is seen at a right angle is a circle.

If $Q$ runs through the semicircle above the fixed diameter, which contains $R$, then $P$ runs once through the Thales circle. If the point $Q$ is on the other semicircle, then the points $P$ lie on the Thales circle drawn over $O R^{\prime}$.

The sought geometric locus is therefore two circles.