Example 3 In the plane where square $ABCD$ is located, find a point $P$ such that $\triangle PAB$, $\triangle PBC$, $\triangle PCD$, and $\triangle PDA$ are all isosceles triangles. Then the number of points $P$ with this property is ( ).
(A) 1
(B) 5
(C) 9
(D) 17
Analysis: For this problem, it is easy to think that there is only 1 point (1 is the intersection of the perpendicular bisectors of the sides, i.e., the center of the regular polygon), this point is the vertex with $P$ as the apex, and $A$, $B$, $C$, $D$ as the base angles of the isosceles triangles. Does there exist an isosceles triangle with $A$, $B$, $C$, $D$ as the vertices of the apex (in this case, the legs of the triangle are the side lengths of the square)? By drawing (as shown in Figure 3), it is not difficult to find that there are $P_{1}$, $P_{2}$, $P_{3}$, $P_{4}$ (inside the square), and $P_{5}$, $P_{6}$, $P_{7}$, $P_{8}$ (outside the square), making a total of 9 points.