For an isothermal reversible compression of an ideal gas

Answer: Option C

Actually the internal energy ($$U$$) of a substance is a function of
$$\begin{array}{rl}& dU=CvdT-\left[P+T\left(\frac{\left(\frac{\partial V}{\partial T}\right)p}{\left(\frac{\partial V}{\partial P}\right)T}\right)dV\right]\\ & \text{For an ideal gas, }PV=RT\\ & \text{So, }\left(\frac{\partial V}{\partial T}\right)p=\frac{R}{P}\text{and }\left(\frac{\partial V}{\partial P}\right)T=-\frac{RT}{P^2}\\ & \text{Hence, }dU=CvdT\end{array}$$
Similarly we can prove $$dH=C_{p}dT$$
So, enthalpy and internal energy are solely functions of temperature in case of ideal gas hence $$\Delta E=\Delta H=0$$