Step 1: Write the line in parametric (distance) form.
A point at signed distance $r$ from $P(2,3)$ along a line at $30^\circ$ to the $x$-axis is $x = 2 + r\cos 30^\circ$, $y = 3 + r\sin 30^\circ$, i.e. $x = 2 + \frac{\sqrt3}{2}r$ and $y = 3 + \frac{1}{2}r$.
Step 2: Substitute into the given curve.
The curve is $x^2 - 2xy - y^2 = 0$. Plug in the parametric expressions; the values of $r$ where the line meets the curve are the roots $r_1, r_2$, and $PA\cdot PB = |r_1 r_2|$.
Step 3: Expand the substitution.
Let $x=2+\frac{\sqrt3}{2}r$, $y=3+\frac12 r$. Then \[ x^2 = 4 + 2\sqrt3\,r + \frac34 r^2,\quad y^2 = 9 + 3r + \frac14 r^2, \] \[ xy = 6 + r + \frac{3\sqrt3}{2}r + \frac{\sqrt3}{4}r^2. \]
Step 4: Collect the $r^2$ and constant terms.
The coefficient of $r^2$ is $\frac34 - 2\cdot\frac{\sqrt3}{4} - \frac14 = \frac12 - \frac{\sqrt3}{2} = \frac{1-\sqrt3}{2}$. The constant term (at $r=0$) is $4 - 2(6) - 9 = -17$.
Step 5: Use the product of roots.
For $A r^2 + B r + C = 0$, the product $r_1 r_2 = \frac{C}{A}$. Here \[ r_1 r_2 = \frac{-17}{\frac{1-\sqrt3}{2}} = \frac{-34}{1-\sqrt3}. \]
Step 6: Rationalise and take the magnitude.
Multiply top and bottom by $1+\sqrt3$: \[ \frac{-34(1+\sqrt3)}{(1-\sqrt3)(1+\sqrt3)} = \frac{-34(1+\sqrt3)}{1-3} = \frac{-34(1+\sqrt3)}{-2} = 17(1+\sqrt3). \] Thus $PA\cdot PB = 17(\sqrt3+1)$.
\[ \boxed{17(\sqrt{3}+1)} \]