5. In the quadrilateral pyramid $P-ABCD$, $BC \parallel AD$, $AD \perp AB$, $AB=2\sqrt{3}$, $AD=6$, $BC=4$, $PA=PB=PD=4\sqrt{3}$. Then the surface area of the circumscribed sphere of the tetrahedron $P-BCD$ is $\qquad$ .
5. $80 \pi$.

As shown in Figure 5, take two trisection points $O_{1}$ and $E$ on $AD$, and connect $BD$, $O_{1}C$, and $CE$. Let $BD$ and $O_{1}C$ intersect at point $H$, and connect $PH$ and $AH$.
Then $AO_{1} = \frac{1}{3} AD = 2 \Rightarrow O_{1}D = BC = 4$.
Since $BC \parallel AD \Rightarrow BC \parallel O_{1}D$
$\Rightarrow$ quadrilateral $BCDO_{1}$ is a parallelogram
$\Rightarrow H$ is the midpoint of $BD$.
Thus, in the right triangle $\triangle ABD$, we have
$AH = BH = DH = \frac{1}{2} BD$
$= \frac{1}{2} \times \sqrt{12 + 36} = 2\sqrt{3}$.
By the Pythagorean theorem,
$O_{1}B = \sqrt{AO_{1}^2 + AB^2} = 4$.
Thus, $O_{1}B = O_{1}D$.
In the right triangle $\triangle O_{1}AB$, we have
$\tan \angle AO_{1}B = \frac{AB}{AO_{1}} = \sqrt{3}$
$\Rightarrow \angle AO_{1}B = \frac{\pi}{3}$.
Since $BC \parallel AD$, we have $\angle CBO_{1} = \frac{\pi}{3}$.
By $BC = O_{1}D = O_{1}B$
$\Rightarrow \triangle O_{1}BC$ is an equilateral triangle
$\Rightarrow O_{1}C = O_{1}B = O_{1}D = 4$
$\Rightarrow O_{1}$ is the circumcenter of $\triangle BCD$.
Given $PA = PB = PD = 4\sqrt{3}$, and $H$ is the midpoint of $BD$, we know $PH \perp BD$.
By $PA = PB$, $AH = BH$, and $PH = PH$
$\Rightarrow \triangle PAH \cong \triangle PBH$
$\Rightarrow \angle PHA = \angle PHB = \frac{\pi}{2}$
$\Rightarrow PH \perp AH$.
Since $PH \perp BD$ and $AH$ intersects $BD$ at point $H$, we have
$PH \perp$ plane $ABCD$,

and $PH = \sqrt{PA^2 - AH^2} = 6$.
Let $O$ be the center of the circumscribed sphere of the tetrahedron $P-BCD$, and connect $OO_{1}$, $OP$, and $OD$. Draw $OF \perp PH$ from point $O$, with the foot of the perpendicular at $F$.
Then the radius $R$ of the circumscribed sphere satisfies
$$
\begin{array}{l}
R^2 = OO_{1}^2 + 4^2 = (6 - OO_{1})^2 + O_{1}H^2 \\
\Rightarrow OO_{1}^2 + 16 = (6 - OO_{1})^2 + 4 \\
\Rightarrow OO_{1} = 2 \\
\Rightarrow R^2 = OO_{1}^2 + 4^2 = 20 .
\end{array}
$$

Thus, the surface area of the circumscribed sphere of the tetrahedron $P-BCD$ is
$$
4 \pi R^2 = 80 \pi
$$