Vapor pressure vs. ambient pressure
The way to understand the effect of the pressure of an inert gas (oxygen here assumed to be inert) on the vapor pressure of a liquid is to consider the effect of the extra pressure on the chemical potential of the liquid: $$\left(\frac{\partial \mu}{\partial P} \right)_T=V_m$$

The additional pressure compresses the liquid and thereby increases its chemical potential, resulting in an increased vapor pressure. However, since $V_{m,~l}$ is relatively small,  the effect on the vapor pressure is also small except at very high pressures. The effect of air at atmospheric pressure and $\pu{298 K}$ on the vapour pressure of water ($\pu{3.2 kPa}$ in the absence of inert gas) is negligible, an increase of $\pu{2.3 Pa}$. 

So $P_A \neq P_B$, but they become nearly identical when $PV_{m,l}\ll RT$.