Example 17. Given as shown, in quadrilateral $ABCD$, $AD=DC=1, \angle DAB=$ $\angle DCB=90^{\circ}, BC, AD$ extended intersect at $P$. Find the minimum value of $AB \cdot S_{\triangle PAB}$.
(1994, Sichuan Province Junior High School Mathematics League Competition)
Let $DP = x$, then $PC = \sqrt{x^2 - 1}$.
Since $\triangle PCD \sim \triangle PAB$,
$\therefore CD \perp AB = PC : PA$.
$\therefore AB = \frac{CD \cdot PA}{PC} = \frac{x + 1}{\sqrt{x^2 - 1}}$.
Let $y = AB \cdot S_{\triangle PAB}$, then
$$
y = \frac{1}{2} AB^2 \cdot PA = \frac{(x + 1)^3}{2(x^2 - 1)}.
$$

Eliminating the denominator and simplifying, we get
$$
x^2 - 2(1 - y)x + 1 + 2y = 0.
$$

Since $x$ is a real number, this equation has real roots, so
$$
\Delta = 4(1 - y)^2 - 4(1 + 2y) = 4y(y - 4) \geqslant 0.
$$

Since $y > 0$, solving gives $y \geqslant 4$,
i.e., the minimum value of $AB \cdot S_{\triangle PAB}$ is 4. E.