6. 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
6. (D).

As shown in the figure, point $P$ can only be on line $l_{1}$ (line BE) and line $l_{2}$, where the distance between $l_{2}$ and line $CD$ is the same as the distance between $l_{1}$ and line $(D)$. Therefore, in isosceles $\triangle PAB$, when $AB$ is the base, the perpendicular bisector of $AB$ intersects $l_{1}$ at $P_{1}$ and $l_{2}$ at $P_{2}$.

In isosceles $\triangle PAB$, when $PA$ is the base, $Y$ is the circumcenter. Therefore, in $\triangle PAB$, when $PB$ is the base, point $E$ meets the condition.
Thus, there are five points that meet the condition: $P_{1}$, $P_{2}$, $P_{3}$, $P_{4}$, and $E$.