The energies of activation for forward and reverse reactions for $A_2+B_2\rightleftharpoons 2AB$ are $180{\text{kJ mol}}^{-1}$ and $200{\text{kJ mol}}^{-1}$ respectively. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by $100{\text{kJ mol}}^{-1}$. The enthalpy change of the reaction ($A_2+B_2\to 2AB$) in the presence of catalyst will be (in ${\text{kJ mol}}^{-1}$):

The enthalpy change of a reaction ($\Delta H$) is the difference between the activation energy of the forward reaction ($E_{a,f}$) and the activation energy of the backward reaction ($E_{a,b}$).
$$\Delta H=E_{a,f}-E_{a,b}$$
In the absence of a catalyst:
$$\Delta H=180-200=-20{\text{kJ mol}}^{-1}$$
A catalyst lowers the activation energy of both forward and backward reactions by the same amount. Let the lowering be $x=100{\text{kJ mol}}^{-1}$.New $E_{a,f}^{\prime }=180-100=80{\text{kJ mol}}^{-1}$ New $E_{a,b}^{\prime }=200-100=100{\text{kJ mol}}^{-1}$ Enthalpy change in the presence of catalyst:
$$\Delta H^{\prime }=E_{a,f}^{\prime }-E_{a,b}^{\prime }=80-100=-20{\text{kJ mol}}^{-1}$$
Since $\Delta H$ is a state function, it does not depend on the path of the reaction or the presence of a catalyst. Thus, it remains $-20{\text{kJ mol}}^{-1}$.