Example 3 In the plane of the regular pentagon $A B C D E$, there exists a point $P$ such that the area of $\triangle P C D$ is equal to the area of $\triangle B C D$, and $\triangle A B P$ is an isosceles triangle. The number of different points $P$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 5
Solution: As shown in Figure 3, point $P$ can only be on line $l_{1}$ (i.e., line $B E$) and line $l_{2}$, 
where the distance between $l_{2}$ and line $C D$ is equal to the distance between $l_{1}$ and line $C D$. Therefore, in isosceles $\triangle P A B$:
(1) When $A B$ is the base, the perpendicular bisector of $A B$ intersects $l_{1}$ at $P_{1}$ and $l_{2}$ at $P_{2}$, so $P_{1}$ and $P_{2}$ meet the conditions.
(2) When $P A$ is the base, with $B$ as the center and $B A$ as the radius, points $P_{3}$ and $P_{4}$ can be intercepted on line $l_{1}$, so $P_{3}$ and $P_{4}$ meet the conditions.
(3) When $P B$ is the base, point $E$ meets the conditions.
Therefore, there are five points $P_{1}$, $P_{2}$, $P_{3}$, $P_{4}$, and $E$ that meet the conditions.
The answer is (D).