Not so sure it is still useful after 2 years, anyway
1) first you need to compute the Laplace transform of the hitting time of llevel $a$ for a standard Brownian Motion, say $B_t$, under a measure $P$.
Define $\tilde{\tau}:= \inf \{t:\ B_t\geq a \}$. In order to do so, we use the hint you proposed, that is
$$E_P(e^{\theta B_t- \frac{\theta^2t}{2}})=1,$$
which entails that, using Doob's Optional Stopping Theorem, since $B_{\tilde{\tau}}=a$,
$$E_P(e^{\theta a- \frac{\theta^2\tilde{\tau}}{2}})=1,$$
or equivalently:
$$e^{-\theta a}= E_P( e^{\frac{\theta^2\tilde{\tau}}{2}}).$$
Now define $\lambda:=\frac{\theta^2}{2} $, so that $\theta = \sqrt{2 \lambda}$.
Replacing this quantities in the above equation, it follows that
$$ E_P( e^{-\lambda\tilde{\tau}})= e^{\sqrt{2 \lambda}}.$$
2) Now we use Girsanov's Theorem. In particular there exists a measure $Q_t$ for which $\frac{dQ_t}{dP_t} = e^{ \mu B_t+ \frac{\mu^2}{2} t}$ and $ \tilde{B}_t =B_t - \mu t $ is a Brownian motion under $Q_t$.
We can apply the result in step 1) for $\tilde{B}_t$, in particular:
$$E_P( e^{-\lambda\tilde{\tau}})= E_Q( e^{-\lambda \tau} e^{ \mu B_\tau+ \frac{\mu^2}{2} \tau} ) = e^{\sqrt{2 \lambda}}.$$
Therefore
$$E_Q(e^{-\tau(-\lambda+\frac{\mu^2}{2} )})= e^{-a(\mu+\sqrt{2\lambda})}.$$
Define $\hat{\lambda}:=-\lambda+\frac{\mu^2}{2}, $ so $\lambda = \tilde{\lambda}+\frac{\mu^2}{2}$. Replacing in the above equation this quantity you get the conclusion.
