= Solution
Let $X_s=W_s-as$ and $M_t=\sup_{0\leq s\leq t}X_s$. Split the crossing event according to its endpoint:
$$
\mathbb P(M_t>b)
=\mathbb P(X_t>b)
+\mathbb P(M_t>b,X_t\leq b).
$$
Under the <Cameron-Martin theorem for a linear drift>, the law of $X$ relative to standard Wiener measure has endpoint density
$$
\exp\!\left(-aW_t-\frac12a^2t\right).
$$
On paths that cross $b$ and end at $x\leq b$, the <Brownian reflection principle> reflects the path after its first hit of $b$ and changes the endpoint to $2b-x\geq b$. Under this reflection the density becomes
$$
e^{-2ab}
\exp\!\left(aW_t-\frac12a^2t\right).
$$
The remaining exponential is the Cameron--Martin density for drift $+a$. Therefore
$$
\mathbb P(M_t>b,X_t\leq b)
=e^{-2ab}\mathbb P(W_t+at\geq b)
=e^{-2ab}\mathbb P(W_t-at\leq-b),
$$
where the last equality uses the symmetry of the <normal distribution>. Subtracting the crossing probability from one gives
$$
\boxed{
\mathbb P\!\left(\sup_{0\leq s\leq t}(W_s-as)\leq b\right)
=\mathbb P(W_t-at\leq b)
-e^{-2ab}\mathbb P(W_t-at\leq-b).}
$$
Equivalently, in terms of the <standard normal distribution function>,
$$
\boxed{
\Phi\!\left(\frac{b+at}{\sqrt t}\right)
-e^{-2ab}\Phi\!\left(\frac{at-b}{\sqrt t}\right).}
$$
This is the <finite-horizon maximum of Brownian motion with negative drift> formula.
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