This is perhaps similar to what mbq meant, but I will elaborate.
The T-p phase diagram of water tells us, for a given temperature and pressure, what phase we will get if we have a bunch of that substance. If I apply different pressures to a bottle of water, I am moving around in the p-direction of the T-p plane. I am not changing the pressure of the triple point of water, just changing the pressure of that particular bottle of water! Similarly, if a tank of water is in a gravitational field, it affects the pressure. In fact, it leads to different pressures at different locations of the tank. It could lead some parts of the tank to freeze, for example. But it does not in any way change the triple point of water itself, which is an intrinsic property of that substance. So I would say that the question is ill-posed. It might be better to ask: what will happen to a tank of water at a given temperature and density if we now apply a gravitational field?
Single species
At the surface of a liquid, molecules are constantly being ejected from the liquid surface and gas molecules rejoining it. When those two rates are equal there's no net change and the system is in equilibrium.
The rate at which liquid water will eject a molecule is roughly independent of pressure and is based solely on temperature. The temperature determines how much kinetic energy the molecules have while the pressure just determines how tightly packed they are and they're already very tightly packed so the density doesn't change much. There are always water molecules right on the surface because of this high density.
The rate at which gaseous water molecules rejoin the water depends on how often one hits the surface which depends on the density. By the ideal gas law, the density is related to pressure and temperature, so if we've already set the temperature, it only depends on pressure. Thus the higher the pressure, the higher the density, the more gaseous water molecules there are to hit the surface and join it.
Locally, the density and pressure of the gaseous water molecules will increase until the rate of rejoining is equal to the rate of evaporation.
Multi gas species
Other gasses can feel free to bombard the surface of the water, compressing it a little bit, and increasing the pressure significantly. However, this doesn't increase the rate at which water molecules escape, so the vapor will still equilibrate at the same partial pressure.
Multi liquid species
This is where Raoult's law comes in. On the liquid surface there are now two species of molecules but the rate at which molecules are ejected is still the same. Now those ejections must be split between species, and according to Raoult's law, it's based on the concentration. If 30% of the liquid's molecules are water then 30% of the ejections will be water, so the rate of ejection is reduced to 30%. So in equilibrium the rate of rejoining must also be 30% so there must be 30% as many collisions thus 30% the density and thus 30% the vapor pressure.
Non Equilibrium
Examining the gas directly above the surface would show a partial pressure of water that approaches the vapor pressure of water for that temperature very quickly. These water molecules would then diffuse through the nitrogen which would slowly lower the local partial pressure if it was not being replenished by additional evaporation from the surface. This evaporation requires energy to overcome the latent heat and as such it will lower the local temperature of the water. Water is a much better heat conductor than air so this heat is drawn from the water.
In some evaporation processes the slow diffusion of the gas dominates, in others, the temperature drops until the diffusion process dominates. This is why liquid nitrogen gets and stays so cold, and why wet bulb temperatures are lower than dry bulb temperatures.
If you'd like to learn more about the steady state solution I would recommend reading about relative/absolute humidity, condensation etc.
If you'd like to learn more about the rate limiting effect look into gas diffusion.
If you'd like to learn more about the evaporation condensation process learning about statistical thermodynamics would help.
Best Answer
The pressure we are talking about is the pressure on (and inside) the water. Imagine pure water in a can; squeeze the can and you increase the pressure. This is the pressure we are talking about.
Neither nor. If other liquids/gases/substances are present (if you have a mixture), then you basically have another material. Then this phase diagram doesn't apply. Another material has another phase diagram.
Exactly. For instance, try googling the phase diagram for salt water and you will see a very different diagram depending on the amount of salt.
I am guessing your hope was that you could separate a mixture into its constituents and use the phase diagrams of each constituent. But that is unfortunately not the case.
A mixture does not behave as two separate materials that just happen to be located on top of each other. That would ignore the interaction between them, which in general severally interferes with their combined phase diagram and with their combined properties. That is why for example the tensile strength of a cobber-nickel alloy is larger than the tensile strength of either pure cobber or pure nickel rather than a value in between.