Okay so, I was waiting for this.
Actually you both are wrong 
See, the question is not about the pressure exerted by a “gas”.
There is no mention of a gas in the question. Water is a liquid at 25⁰C.
The question actually asks about “Vapour Pressure”.
Vapour Pressure is the Pressure exerted by vapours of a liquid (vapours of the liquid exist in equilibrium) on the surface of that liquid.
Thus, liquid
vapour (equilibrium), ∆H=+ve.
Enthalpy is positive since energy is required for converting liquid to vapour.
Kp (equilibrium constant)= Pvap/Pliq.
Pressure exerted by a pure liquid is 1 (constant) wrt to Pressure of the vapours.
Thus, Pvap or vapour pressure= Kp.
And Kp doesn’t depend upon Volume, surface, amount of liquid, etc.
Kp only depends upon Temperature and nature of the liquid.
Thus, Vapour Pressure of a liquid only depends on Temperature and Nature of liquid, generally related as
Log(P2/P1)=∆Hvap((1/T1)-(1/T2))/2.303R
Thus, Ans will be none of these, (option 4) as the vapour pressure won’t change, because the temperature doesn’t change.
You can also Google this, it’s pretty interesting and surprising at the same time!