3. (C)

Solution Let $A P=x$, then $P B=7-x$,
(1) If $\triangle P A D \backsim \triangle P B C$, then $\frac{x}{7-x}=\frac{2}{3}, \quad \therefore x=\frac{14}{5}<7$. This meets the condition.
(2) If $\triangle P A D \backsim \triangle C B P$, then
$$
\frac{x}{3}=\frac{2}{7-x},
$$
$\therefore x_{1}=1, x_{2}=6$ also meet the condition. Therefore, there are 3 points $P$ that satisfy the condition.
3. As shown in the figure, in the right trapezoid $A B C D$, $A B=7$. $A D=2, B C=3$, if point $P$ on side $A B$ makes the triangle with vertices $P, A, D$ similar to the triangle with vertices $P, B, C$, then the number of such points $P$ is
(A) 1;
(B) 2;
(C) 3;
(D) 4.