= Solution
Fix $\lambda>0$ and choose the positive root
$$
\theta=b+\sqrt{b^2+2\lambda}>0,\qquad \frac{\theta^2}{2}-b\theta=\lambda.
$$
Use the <Exponential martingale for Brownian motion> $M_t=\exp(\theta B_t-\theta^2t/2)$. Before the linear-boundary <stopping time> $\tau$, path continuity gives $B_s<a+bs$. At the boundary there is equality. Consequently
$$
0\leq M_{t\wedge\tau}\leq e^{\theta a}e^{-\lambda(t\wedge\tau)}\leq e^{\theta a}.
$$
The <optional stopping theorem> at the bounded <stopping time> $t\wedge\tau$ yields $\mathbb E M_{t\wedge\tau}=1$. Separate the two events:
$$
1=e^{\theta a}\mathbb E[e^{-\lambda\tau}\mathbf1_{\{\tau\leq t\}}]+\mathbb E[M_t\mathbf1_{\{\tau>t\}}].
$$
The last term is at most $e^{\theta a-\lambda t}$ and tends to zero. The first term converges by the <monotone convergence theorem>. Taking $e^{-\lambda\infty}=0$ therefore proves
$$
\boxed{\mathbb E e^{-\lambda\tau}=\exp\!\left[-a\left(b+\sqrt{b^2+2\lambda}\right)\right].}
$$
This bounded-stopping argument does not assume in advance that $\tau$ is finite. For $b\leq0$, letting $\lambda\downarrow0$ makes the right side tend to one, so it also proves $\mathbb P(\tau<\infty)=1$. Setting $b=0$ gives the <Brownian first-passage Laplace transform>
$$
\boxed{\varphi_a(\lambda)=e^{-a\sqrt{2\lambda}},\qquad\lambda\geq0,\qquad C=\sqrt2.}
$$
The zero-parameter value follows from the just-proved almost-sure finiteness.
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