Let $P(x_0, x_0 + \frac{2}{x_0})$. The expression for line $PA$ is then
$$y - \left(x_0 + \frac{2}{x_0}\right) = -(x - x_0),$$
or
$$y = -x + 2x_0 + \frac{2}{x_0}.$$
From
$$\begin{cases} y = x, \\ y = -x + 2x_0 + \frac{2}{x_0}, \end{cases}$$
we get $A(x_0 + \frac{1}{x_0}, x_0 + \frac{1}{x_0})$.
On the other hand, we have $B(0, x_0 + \frac{2}{x_0})$. Then $\vec{PA} = (\frac{1}{x_0}, -\frac{1}{x_0})$ and $\vec{PB} = (-x_0, 0)$. Therefore,
$$\vec{PA} \cdot \vec{PB} = \frac{1}{x_0} \cdot (-x_0) = -1.$$
The answer is $-1$.

As seen in Fig. 1.1, the distances from $P(x_0, x_0 + \frac{2}{x_0})$ to lines $y = x$ and $y$-axis, respectively, are

Fig. 1.1
$$|PA| = \frac{|x_0 - \left(x_0 + \frac{2}{x_0}\right)|}{\sqrt{2}} = \frac{\sqrt{2}}{x_0}$$
and
$$|PB| = x_0.$$
Since $O$, $A$, $P$ and $B$ are concyclic points, then
$$\angle APB = \pi - \angle AOB = \frac{3\pi}{4}.$$
Therefore, $\overrightarrow{PA} \cdot \overrightarrow{PB} = |\overrightarrow{PA}| \cdot |\overrightarrow{PB}| \cdot \cos \frac{3\pi}{4} = -1$. $\square$