Let $ABCD$ be the given quadrilateral and $P, Q$ be the given points. For the sake of contradiction assume that the lines $AB$ and $CD$ intersect at a point $E$. By the given condition there exist the isogonal conjugates $P_1, Q_1$ of $P, Q$ wrt $ABCD$ respectively and moreover the segments $PP_1$ and $QQ_1$ share a common midpoint, i.e. $PQP_1Q_1$ is a parallelogram. By the theorem about isogonals the points $PQ\cap P_1Q_1$ and $PP_1\cap QQ_1$ at infinity are isogonal conjugates wrt $ABCD$ and also wrt $EBC$. But because the isogonal conjugate of the point at infinity wrt a triangle lies on its circumcircle we arrive to a contradiction (the points $PQ\cap P_1Q_1$ and $PP_1\cap QQ_1$ don't lie on the circumcircle of $EBC$). Thus $AB\parallel CD$ and similarly $AD\parallel BC$.

Let $P,Q$ be these two points, $O$ be the center of concentric circles. Existence of pedal circles of $P,Q$ wrt quadrilateral implies existence of isogonal conjugates of $P,Q$ wrt it, and hence exist two conics inscribed in quadrilateral and centered at $O$. As lines through quadrilateral sides are all $4$ common tangents to these conics, we by the symmetry wrt $O$ get the result.