Solution:

A rotation keeps its center equidistant from each point and its image. So $P$ is equidistant from $A$ and $A',$ and from $B$ and $B',$ which puts it at the intersection of two perpendicular bisectors. The bisector of $BB'$ from $(3,3)$ to $(4,3)$ is $x = 3.5.$ The bisector of $AA'$ from $(1,2)$ to $(3,1)$ is $2x - y = 2.5.$ Then $y = 2(3.5) - 2.5 = 4.5,$ so $P = (3.5, 4.5)$ and $|r - s| = |3.5 - 4.5| = 1.$ Thus, E is the correct answer.