Define the Grassman variables $\sqrt{2}\eta=\psi+\phi$ and $\sqrt{2}\chi=\psi-\phi$.
Then $$ \eta^{T} A \chi = {1\over 2} (\psi+\phi)^{T} A (\psi - \phi) = {1\over 2}(\psi^T A\psi - \phi^T A \phi)$$
Since the middle terms cancel by antisymmetry of A and the Grassmann property. The result proves the equality of the two actions, so that the determinant of A is equal to the product of Pfaffians.
This is not quite what you wanted--- there is a minus sign in the above. But the minus sign only gives an overall minus to the whole expression if the dimension of A is odd, and it is necessary.
