16.7 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.
[Solution 1] Let $AP = x$, then $PB = 7 - x$. (As shown in Figure (1)).
(1) If $\triangle PAD \sim \triangle PBC$, then $\frac{x}{7-x} = \frac{2}{3}$,
$\therefore x = \frac{14}{5}\frac{5}{2}\right)$, a circle with radius $\frac{5}{2}$ must intersect line $AB$ at two points, denoted as $P_{2}$ and $P_{3}$.

From
$$
\begin{aligned}
EP_{2} & = EP_{3} = \sqrt{OP_{2}^2 - OE^2} \\
& = \sqrt{\left(\frac{5\sqrt{2}}{2}\right)^2 - \left(\frac{5}{2}\right)^2} \\
& = \frac{5}{2} < \frac{7}{2} = EA.
\end{aligned}
$$

It follows that $P_{2}$ and $P_{3}$ lie on segment $AB$.
It is easy to prove that $\triangle P_{2}AD \sim \triangle CBP_{2}$, $\triangle P_{3}AD \sim \triangle CBP_{3}$.
Therefore, $P_{2}$ and $P_{3}$ also meet the conditions.
The following proof shows that $P_{1}$ must not coincide with $P_{2}$ (similarly, $P_{3}$).
If not, there should be $\angle BP_{1}C = \angle AP_{1}D = \angle BCP_{1} = 45^{\circ}$.
$\triangle P_{1}AD$ and $\triangle P_{1}BC$ are both isosceles right triangles.
And $AB = AP_{1} + P_{1}B = AD + BC = 2 + 3 = 5$.
This contradicts the condition $AB = 7$.
In summary, there are three points $P$ that meet the conditions.
Therefore, the answer is $(C)$.