A straightforward approach is to simply integrate the density of $X_t$ at time $a$ (which will be normally distributed with mean $\mu$ and variance $\sigma^2 a$) against the probability of hitting 0 conditional on the value at time $a$ (which is also known in closed-form). This will give you a messy integral (with an exponential multiplied by a cumulative-normal) but it should be reducible to a (messy sum of) bivariate cumulative normal(s).

The value we want to compute is
$$\int_0^\infty \mathbb{P}[X_\xi>0 \text{ for } a\leq\xi\leq T\ |\ X_a=z] e^{-z^2/2\sigma^2T}\frac{dz}{\sigma\sqrt{T}\sqrt{2\pi}}$$
where I'm integrating the density at time $a$ for positive values against the non-hitting time.

The next step is to observe that the probability
$\mathbb{P}[X_\xi>0 \text{ for } a\leq\xi\leq T\ |\ X_a=z]$
is equal to the probability
$\mathbb{P}[X_\xi>-z \text{ for } 0\leq\xi\leq T-a]$ but this probability is equal to a difference of (basically) cumulative normals (it's just a hitting time computation for a (scaled) Brownian motion with drift). Then plug that formula into the above integral.

A quick calculation (might be wrong, so beware) gives me
$$\mathbb{P}[X_\xi>-z \text{ for } 0\leq\xi\leq T-a] =
\Phi\left[\frac{-z+\alpha (T-a)}{\sigma\sqrt{T-a}}\right]
- e^{2\alpha z/\sigma^2}\Phi\left[\frac{z+\alpha (T-a)}{\sigma\sqrt{T-a}}\right] $$
where $\Phi[z]=\int_{-\infty}^z e^{-\xi^2/2}\frac{d\xi}{\sqrt{2\pi}}$ is the standard cumulative normal distribution function. (This follows from application of Girsanov to a reflection argument, a well-known result.)