A reaction has a forward activation energy of 94 kJ/mol and Δ H = -38 kJ/mol…

Step 1. Without the catalyst. $E_a' = 94 - (-38) = 132\ \text{kJ/mol}$.

Step 2. With the catalyst. $E_a' = 61 - (-38) = 99\ \text{kJ/mol}$. The reverse barrier fell by $33\ \text{kJ/mol}$, the same as the forward barrier.

Step 3. The enthalpy change. The reactants and the products are the same, so $\Delta H$ is still $-38\ \text{kJ/mol}$.

Step 4. The equilibrium. The forward and the reverse reactions are speeded up by the same factor. So the equilibrium constant does not change.