H2Andy:well, that's not the scenario presented
the scenario presented was a flooded tube
as for the "air pocket" scenario, an air pocket could only exist at 1 ATM or "above" 1 ATM. at 1.1ATM, the water would "push up" the air back to 1 ATM.
In an air pocket scenario, the air pressure in the pocket will match the water pressure at that depth. It will never be 1atm unless the pocket is at sea level. In the original question the cave at point C, 33 feet down, will be 2atm.
so, even with an air pocket above you, you still calculate as i indicated:
[water depth/33] + 1 = ATM
Assuming you mean the depth from the surface exposed to outside air and not the depth from the surface inside the air pocket, then that's correct.