8. In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, $A B=$ $A D=5, A A_{1}=1, M$ is a point on edge $A B$ and satisfies $3 \overrightarrow{A M}+2 \overrightarrow{B M}=\mathbf{0}$. A point $P$ is randomly selected within the square base $A B C D$, and the distance from point $P$ to line $A_{1} D_{1}$ is $d_{1}, P M=d_{2}$. Then the probability that the point $P$ satisfies $d_{1}^{2}-d_{2}^{2} \geqslant 1$ is $\qquad$.
8. $\frac{32}{75}$.

As shown in Figure 2, let point $P$ have projections $R$ and $Q$ on $AD$ and $A_1D_1$, respectively.
Then $\triangle PQR$ is a right

triangle,
$$
\begin{array}{l}
\angle PRQ=90^{\circ}, \\
RQ=1 .
\end{array}
$$

From $1 \leqslant d_1^2 - d_2^2$
$$
\begin{array}{l}
= PQ^2 - PM^2 = (PR^2 + RQ^2) - PM^2 \\
\Rightarrow PR \geqslant PM .
\end{array}
$$

According to the definition of a parabola, the locus of point $P$ that satisfies $PR = PM$ is a segment of a parabola.

Take the midpoint $P_1$ of segment $AM$, and take point $P_2$ on edge $BC$ such that $BP_2 = 4$. It is easy to see that the parabola passes through points $P_1$ and $P_2$.

Using analytic geometry, the area of the parabola with focus $M$ and directrix $AD$ within the square $ABCD$ is $\frac{32}{3}$.
Therefore, the required probability is $\frac{32}{3} \div 25 = \frac{32}{75}$.