The ionisation potential of $X_{(g)}^{-}$ is numerically equal to :

$X^{-}\to X+e^{-};$  $\Delta H=I.Pof{X}^{-}$
$X+e^{-}\to X^{-};$ $\Delta H=-E.AofX$ 

Comparing the above two equations we get :
  $I.Pof{X}^{-}=E.AofX$