1958 AMC 12 Problem 47

Because $PQ\parallel BD$ and $PS\parallel AF,$ identify the small parallelogram near $F$ and $S$

Use the intersection $T=PQ\cap AC$ to compare the right triangles $PTR$ and $ATQ$

Since $AF\perp BD$ and $PQ\perp AF,$ we have $PQ\parallel BD.$ Also $PS\perp BD,$ so $PS\parallel AF.$ Thus quadrilateral $QPSF$ is a rectangle, and $$PS=QF.$$ Let $T=PQ\cap AC.$ The diagonals of a rectangle make equal angles with side $AB,$ so $\angle PAT=\angle APT,$ giving $AT=PT.$ The right triangles $ATQ$ and $PTR$ are similar because they share the angle at $T.$ Since their hypotenuses $AT$ and $PT$ are equal, $AQ=PR.$ Therefore $$PR+PS=AQ+QF=AF.$$

Thus, the correct answer is D.