16. As shown in Figure 2, two semicircles with centers at $A$ and $B$ and radii of 2 and 1, respectively, are internally tangent to the semicircle with diameter $JK$, and the two smaller semicircles are also tangent to each other. $\odot P$ is externally tangent to both smaller semicircles and internally tangent to the larger semicircle $\Gamma$. Then the radius of $\odot P$ is ( ).
(A) $\frac{3}{4}$
(B) $\frac{6}{7}$
(C) $\frac{\sqrt{3}}{2}$
(D) $\frac{5 \sqrt{2}}{8}$
(E) $\frac{11}{12}$
16. B.

Let the center of circle $\Gamma$ be $C$, and connect $P A, P C, P B$. In $\triangle A B P$, by Stewart's Theorem, we have
$$
\begin{array}{c}
A B \cdot A C \cdot B C+A B \cdot C P^{2} \\
=A C \cdot B P^{2}+B C \cdot A P^{2} .
\end{array}
$$

From the given tangency relations, we get
$$
\begin{array}{l}
A B=3, A C=1, B C=2, \\
A P=2+r, B P=1+r, C P=3-r,
\end{array}
$$

where $r$ is the radius of $\odot P$.
Substituting into equation (1), we get
$$
\begin{array}{l}
3 \times 1 \times 2+3(3-r)^{2}=(1+r)^{2}+2(2+r)^{2} \\
\Rightarrow r=\frac{6}{7} .
\end{array}
$$