Lets assume $Z$ is a random variable defined over a support of $\mathbb{R}$ and $X_1,...,X_n$ are i.i.d and have cdf $\Phi$, in which case
\begin{align}
\text{Prob}(a)
&\equiv \mathbb{P}(X_1 \le Z + a, X_2 \le Z + a, ..., X_n \le Z + a) \\[18pt]
&= \mathbb{E}_Z[\mathbb{P}(X_1 \le Z + a, X_2 \le Z+ a, ..., X_n \le Z + a | Z = z)] \\[12pt]
&= \int \limits_{-\infty}^{\infty} \mathbb{P}(X_1 \le Z + a, X_2 \le Z+ a, ..., X_n \le Z + a | Z = z) \phi(z) \,dz \\[6pt]
&= \int \limits_{-\infty}^{\infty}\mathbb{P}(X_1 \le z + a, X_2 \le z + a, ..., X_n \le z + a) \phi(z) \,dz \\[6pt]
&= \int \limits_{-\infty}^{\infty}\mathbb{P}(X_1 \le z + a) \cdot \mathbb{P}(X_2 \le z + a) \cdots \mathbb{P}(X_n \le z + a) \phi(z) \,dz \\[6pt]
&= \int \limits_{-\infty}^{\infty}\Phi(z + a)^n \phi(z) \,dz. \\[6pt]
\end{align}
Intuitively, $z$ (the variable of integration) is a scalar as opposed to a vector because it is the same variable that all $X_1,...,X_n$ are being compared with, and it allows us to compute the total probability utilizing the conditional probability given $Z=z$, which is much easier probability to compute. This alternative, easier formulation is allowed as per the law of total probability.
Observe that the integration variable would instead be the vector $(z_1,...,z_n$) if we were instead computing $\mathbb{P}(X_1 \le Z_1 + a, X_2 \le Z_2 + a, ..., X_n \le Z_n + a.)$