The most relevant thing I could find is the paper An explicit expression for the distribution of the supremum of brownian motion with a change point by Benzion Boukai. The relevant result is in Lemma 2 in section 2.

(If you can't access that paper, then the technical report has the same result in Theorem 1.) I reproduce the result here.

Let $B_{t_0}^{(x,y)}(t) $ be the Brownian bridge with $B^{(x,y)}_{t_0}(0) = x$ and $B^{(x,y)}_{t_0}(t_0) = y$. Then

$$\Pr\left(\sup_{0 < t\leq t_0}B_{t_0}^{(x,y)}(t) > z \right) = \begin{cases} e^{-2(z-x)(z-y)/t_0} & z > \max(x,y) \\ 1 & \text{otherwise}\end{cases}\,.$$

In your case, since $x = y = a$,

$$\Pr\left(\sup_{0 < t\leq t_0}B_{t_0}^{(a,a)}(t) > z \right) = \begin{cases} e^{-2(z-a)^2/t_0} & z > a \\ 1 & \text{otherwise}\end{cases}\,.$$

The most relevant thing I could find is the paper An explicit expression for the distribution of the supremum of brownian motion with a change point by Benzion Boukai. The relevant result is in Lemma 2 in Section 2.

(If you can't access that paper, then the technical report has the same result in Theorem 1.) I reproduce the result here.

Let $B_{t_0}^{(x,y)}(t) $ be the Brownian bridge with $B^{(x,y)}_{t_0}(0) = x$ and $B^{(x,y)}_{t_0}(t_0) = y$. Then

$$\Pr\left(\sup_{0 < t\leq t_0}B_{t_0}^{(x,y)}(t) > z \right) = \begin{cases} e^{-2(z-x)(z-y)/t_0} & z > \max(x,y) \\ 1 & \text{otherwise}\end{cases}\,.$$

In your case, since $x = y = a$,

$$\Pr\left(\sup_{0 < t\leq t_0}B_{t_0}^{(a,a)}(t) > z \right) = \begin{cases} e^{-2(z-a)^2/t_0} & z > a \\ 1 & \text{otherwise}\end{cases}\,.$$