= Solution
Use the time-inverted Brownian motion $B_u=uW_{1/u}$. For $s>0$, set $u=1/s$. Then
$$
W_s-as\leq b
\quad\Longleftrightarrow\quad
\frac{W_s}{s}-a\leq\frac bs
\quad\Longleftrightarrow\quad
B_u-bu\leq a.
$$
As $0<s\leq t$ corresponds exactly to $u\geq1/t$, and the inequality at $s=0$ is automatic because $b>0$, this is the pathwise identity
$$
\left\{\sup_{0\leq s\leq t}(W_s-as)\leq b\right\}
=\left\{\sup_{u\geq1/t}(B_u-bu)\leq a\right\}.
$$
Since $B$ and $W$ have the same law by <time inversion of Brownian motion>,
$$
\boxed{
\mathbb P\!\left(\sup_{0\leq s\leq t}(W_s-as)\leq b\right)
=\mathbb P\!\left(\sup_{u\geq1/t}(W_u-bu)\leq a\right).}
$$
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