What does the pressure-position graph look like when an upside-down mug with air is pushed underwater? 
I know that the pressure increases linearly between A and B, once the mug is at B there will be a jump in pressure, and then the pressure will be constant between B and C. I'm just not sure if this sudden jump in pressure is increasing or decreasing. I think the pressure would suddenly increase, but I'm not sure (I inserted what I think it looks like below). Could someone please help me reason through it?

 A: The thing that cup is filled with an air will not change anything. Pressure will increase gradually and continuously due to bigger depths and air in the cup will get constantly compressed more and more, due to fact that water pressure will equate to ideal gas pressure in the cup:
$$ \rho g h = \frac {nRT}{V} \tag 1$$
From (1) you can see that as depth increases, volume of gas decreases (it gets compressed by water pressure), assuming isothermal process.
If something would force process to be isochoric (of a constant volume), then along with depth- temperature of gas inside a cup must increase to compensate for water pressure changes. In any case, I do not believe that process will be isobaric between points BC (as drawn in a chart), because for that to happen, while gas volume decreases something must cool the gas to lower its temperature constantly- which is highly unlikely.
A: Your diagram is essentially correct for a stationary mug positioned as shown in the figure (which to me is clearly your intention despite some apparent confusion on that point).
If you consider a line to the side of the mug, one that is only in water, then the pressure simply increases linearly from top to bottom by $\Delta P = \rho g \Delta h$. That determines the pressure in the water at A, C, and D.
The pressure in the air at C is equal to the pressure in the water at C. Otherwise the water and the air would be accelerating.
Because air is not very dense ($\rho \approx 0$) the pressure in the air at B is nearly the same as the pressure in the air at C ($\Delta P=\rho g \Delta h \approx 0$). This difference in pressure between the air inside pushing up on the mug and the water outside pushing down is what causes the mug to be buoyant.
So your graph is essentially correct.
