5. As shown in Figure 1, in the quadrilateral pyramid $P-A B C D$, $A D$ 
$/ / B C, \angle A B C=$ 
$90^{\circ}, P A \perp$ plane 
$A B C D, P A=3, A D=2, A B=2 \sqrt{3}, B C=6$. Then the size of the dihedral angle $P-B D-A$ is
5. $60^{\circ}$.

Draw $A F \perp B D$ at point $F$.
Since $P A \perp$ plane $A B C D$, we know $P F \perp B D$.
Thus, $\angle P F A$ is the size of the dihedral angle $P-B D-A$.
Because $\angle A B C=90^{\circ}, A D / / B C$, so, in the right triangle $\triangle A B D$, given $A B=2 \sqrt{3}, A D=2$, we know $B D=4, A F=\sqrt{3}$.
Therefore, $\tan \angle P F A=\frac{P A}{A F}=\frac{3}{\sqrt{3}}=\sqrt{3}$
$$
\Rightarrow \angle P F A=60^{\circ} \text {. }
$$