The given reaction is $A \rightarrow B$. Enthalpy of reaction, $\Delta H = -4.2 \text{ kJ mol}^{-1}$. The negative sign indicates that the reaction is exothermic. Therefore, the potential energy of the product (B) will be lower than the potential energy of the reactant (A) by $4.2 \text{ kJ mol}^{-1}$. Activation energy, $E_a = 9.6 \text{ kJ mol}^{-1}$. This means the energy barrier (peak of the curve) is $9.6 \text{ kJ mol}^{-1}$ above the energy of the reactant (A). The correct potential energy profile will show the reactant A at a certain energy level, a peak $9.6 \text{ kJ mol}^{-1}$ above A, and the product B at an energy level $4.2 \text{ kJ mol}^{-1}$ below A.