14. As shown in Figure 6, it is known that quadrilateral $ABCD$ is inscribed in a circle $\odot O$ with a diameter of 3, diagonal $AC$ is the diameter, the intersection point of diagonals $AC$ and $BD$ is $P$, $AB=BD$, and $PC=0.6$. Find the perimeter of quadrilateral $ABCD$.
$1+$ Let the heart of the shadow be क”, so $\mathrm{BH} / \mathrm{C}$.
From this, $\triangle O P B C \triangle C P D, \frac{C D}{B O}=\frac{C P}{P O}$,

which means $\frac{C D}{1.5}=\frac{0.6}{1.5-0.6}$. Therefore, $C D=1$.
Thus, $A D=\sqrt{A C^{2}-C D^{2}}=\sqrt{9-1}=2 \sqrt{2}$.
Also, $O H=\frac{1}{2} C D=\frac{1}{2}$, so,
$$
\begin{array}{l}
A B=\sqrt{A H^{2}+B H^{2}}=\sqrt{2+4}=\sqrt{6}, \\
B C=\sqrt{A C^{2}-A B^{2}}=\sqrt{9-6}=\sqrt{3} .
\end{array}
$$

Therefore, the perimeter of quadrilateral $A B C D$ is
$$
1+2 \sqrt{2}+\sqrt{3}+\sqrt{6} \text {. }
$$