8. Given a moving point $P(x, y)$ satisfies
$$
\left\{\begin{array}{l}
2 x+y \leqslant 2, \\
x \geqslant 0, \\
\left(x+\sqrt{x^{2}+1}\right)\left(y+\sqrt{y^{2}+1}\right) \geqslant 1 .
\end{array}\right.
$$

Then the area of the figure formed by the moving point $P(x, y)$ is
8. 2 .

From equation (1), we have
$$
\begin{array}{l}
x+\sqrt{x^{2}+1} \geqslant \sqrt{y^{2}+1}-y \\
\Rightarrow \ln \left(x+\sqrt{x^{2}+1}\right) \geqslant \ln \left(\sqrt{y^{2}+1}-y\right) .
\end{array}
$$

It is easy to see that $f(x)=\ln \left(x+\sqrt{x^{2}+1}\right)$ is a strictly increasing function.
$$
\begin{array}{l}
\text { Therefore, equation (1) } \Leftrightarrow f(x) \geqslant f(-y) \\
\Leftrightarrow x \geqslant -y \Leftrightarrow x+y \geqslant 0 .
\end{array}
$$

From $\left\{\begin{array}{l}2 x+y \leqslant 2 \\ x+y \geqslant 0\end{array}\right.$, we get the intersection point coordinates as $(2,-2)$.
Thus, the area of the region formed by the moving point $P$ is
$$
\frac{1}{2} \times 2 \times 2=2
$$