It is well known that The Kolmogorov distribution is the distribution of the random variable $$ {\displaystyle K=\sup _{t\in [0,1]}|B(t)|} $$ where B(t) is the Brownian bridge:
$$ B(t) = (W_t|W_1=0) $$
My question is the following - can we say anything about distribution of $K$ if $B(t)$ is defined as:
$$ B(t) = (W_t|W_1=\textbf{a}) $$
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 $x = y = a$,
$$\Pr\left(\sup_{0 < t\leq t_0}B_{t_0}^{(a,a)}(t) > z \right) = \begin{cases} e^{-2(z-a)^2/t_0} & z > a \\ 1 & \text{otherwise}\end{cases}\,.$$
