For a spontaneous process, free energy

Answer: Option C

From second law of thermodynamics
$$\begin{array}{rl}& TdS⩾\delta Q\\ & \Rightarrow TdS⩾dU+\delta W\end{array}$$
For an irreversible process $$TdS-dU-\delta W>0$$
And for a reversible process $$TdS-dU-\delta W=0$$
For any spontaneous process there should be finite changes so, we can consider it as an irreversible process and we know for irreversible process from second law of thermodynamics by above discussion: $$TdS-dU-PdV>0$$
Under constant temperature and volume process $$-dF>0\Rightarrow dF<0$$
Similarly for an constant temperature and pressure process $$d\left(TS-U-PV\right)>0\Rightarrow dG<0$$