= Solution
For zero drift, the reflection principle gives the known <Brownian first-passage time> density
$$
h_0(t)=\frac{a}{\sqrt{2\pi t^3}}e^{-a^2/(2t)},\qquad t>0.
$$
To obtain a general drift, on a fixed finite horizon use the exponential <martingale> $M_t=e^{cB_t-c^2t/2}$ as change-of-measure density. The <Novikov condition> holds for constant $c$, and the <Girsanov theorem> says that the coordinate process $B$ has drift $c$ under the new measure. Thus the hitting law of $B$ under that measure is the law of $B+ct$ under the original measure. <Conditional expectation> of $M_T$ at a hitting time bounded by $T$ equals its stopped value, by the <martingale> <optional sampling theorem>. At $H_a=t$ this value is $e^{ca-c^2t/2}$. The <drifted Brownian first-passage density> is obtained by multiplying $h_0$, giving \b[$\boxed{h_c(t)=a e^{-(a-ct)^2/(2t)}/\sqrt{2\pi t^3}}$.]
To verify the integrated formula, put $u=(a-cT)/\sqrt T$ and $v=(-a-cT)/\sqrt T$. The identity $e^{2ac}\phi(v)=\phi(u)$ for the standard normal density gives
$$
\frac d{dT}[\overline\Phi(u)+e^{2ac}\Phi(v)]=\phi(u)\frac{a+cT+a-cT}{2T^{3/2}}=h_c(T).
$$
Both terms vanish as $T\downarrow0$, so \b[$\boxed{\mathbb P(H_a\leq T)=\overline\Phi((a-cT)/\sqrt T)+e^{2ac}\Phi((-a-cT)/\sqrt T)}$.] The distribution has total mass one for $c\geq0$ and mass $e^{2ac}$ for $c<0$; in the latter case the remaining probability is an atom at infinity.
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