The correct option is A ΔG>0 and ΔS<0⇒ΔH<0, so it is exothermic
To answer this question, let us recall our fundamental equation
ΔG=ΔH−TΔS
We already know and accept the fact that adsorption is spontaneous. So we can conclude that ΔG<0
Now, when the gases get adsorbed on the surface of the solid from their native state, their entropy reduces. An easy way to think about why this happens is that when the gases were in their native state they had more 'randomness'.
But when they are adsorbed on the surface, they are more restricted and are less 'random'.
So, we can say that entropy change ΔS<0
From the equation, if ΔG<0 and −TΔS>0 since we have negative sign, then we need ΔH<0 and also higher in value than TΔS