= Solution
By continuity of <Brownian motion>, the event that the last strict crossing of $b$ occurs by time $t$ is
$$
\{T\leq t\}
=\left\{\sup_{u\geq t}(W_u-au)\leq b\right\}.
$$
Apply part (b) with $a$ and $b$ interchanged and its time parameter replaced by $1/t$. This turns the right-hand side into
$$
\mathbb P\!\left(
\sup_{0\leq s\leq1/t}(W_s-bs)\leq a
\right).
$$
Part (c), now with drift $b$, barrier $a$, and horizon $1/t$, gives
$$
\boxed{
\mathbb P(T\leq t)
=\Phi\!\left(a\sqrt t+\frac b{\sqrt t}\right)
-e^{-2ab}\Phi\!\left(\frac b{\sqrt t}-a\sqrt t\right),
\qquad t>0.}
$$
This is the distribution function of the <last passage time above a level for Brownian motion with negative drift>. Taking $t\downarrow0$ gives
$$
\mathbb P(T=0)=1-e^{-2ab},
$$
the probability that the negatively drifted Brownian path never exceeds $b$, consistent with the <infinite-horizon crossing probability for Brownian motion with negative drift>.
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