They had to make some hand-waving assumption they called "absence of holonomy" to get rid of the quadratic term in a, which is of course where all the action happens in this duality to give you fermions and everything.

Once its linearized, the path integral is evaluated by analogy with the 1-dimensional integral

$$\int dx e^{i x y} = \delta(y).$$

EDIT: I was being a bit too cavalier and was going to end up with $\delta(\tilde a - A)$, which is too strong. I was altogether dropping the $ada$ term. A better thing to do is complete the square: redefine $a' = a + (1/2)(\tilde a - A)$. Then the integrand becomes

$$ a' da' - (1/4)(\tilde a - A)d(\tilde a - A).$$

One should worry about the factors of 1/2 and 1/4 that we ended up with. In this form, we're free to do the path integral over $a'$, which on a simply connected manifold can be computed as a Gaussian integral and the saddle point (which is unique on a simply connected manifold) is $a' = 0$, in other words $a = A - \tilde a$.