$\mathrm{AB}=\mathrm{AC}$

$\Rightarrow \sqrt{(0-3)^{2}+(2-p)^{2}}=\sqrt{(0-p)^{2}+(2-5)^{2}}$
$\Rightarrow \sqrt{(-3)^{2}+(2-p)^{2}}=\sqrt{(-p)^{2}+(-3)^{2}}$

Squaring both sides, we get

$(-3)^{2}+(2-p)^{2}=(-p)^{2}+(-3)^{2}$

$\Rightarrow 9+4+p^{2}-4 p=p^{2}+9$

$\Rightarrow 4 p=4$

$\Rightarrow \mathrm{p}=1$

$\mathrm{AB}=\sqrt{(0-3)^{2}+(2-\mathrm{p})^{2}}$

$=\sqrt{(-3)^{2}+(2-1)^{2}} \quad(\because p=1)$

$=\sqrt{9+1}$

$=\sqrt{10}$ units

Therefore, $p=1$ and $A B=\sqrt{10}$ units