9. Given that point $P(x, y)$ satisfies $|x|+|y| \leqslant 2$. Then the probability that the distance $d \leqslant 1$ from point $P$ to the $x$-axis is $\qquad$
9. $\frac{3}{4}$.

As shown in Figure 2, the region defined by $|x|+|y| \leqslant 2$ forms a square $ABCD$.
It is easy to see that the area of square $ABCD$ is 8.
Since the line $y=1$ intersects $AD$ and $AB$ at points $E$ and $F$ respectively, and the line $y=-1$ intersects $CD$ and $CB$ at points $H$ and $G$ respectively, we have:
$$
S_{\triangle AEF}=1, S_{\triangle CHG}=1, S_{\text{hexagon } EFBGHD}=6.
$$

Therefore, the probability $p$ of point $P$ is $p=\frac{S_{\text{hexagon } EFBGHD}}{S_{\text{square } ABCD}}=\frac{3}{4}$.