The water inside would be at almost the same pressure as the water outside. The difference in pressure would be due to the residual stresses in the bottle that are resisting being deformed. If it was a very thick-walled vessel similar to a vacuum chamber, then there could be a substantial pressure difference. If the bottle were like a balloon, then there would be essentially no pressure difference. If the bottle were a relatively thin-walled metal cylinder (with some kind of endcaps), it would deform into an oval shape instead of remaining cylindrical. This is a form of buckling. So it would be the bending stiffness that would determine the pressure difference rather than a hoop stress in the wall that you would calculate if you assumed the cylinder remained circular. So the pressure difference would be much smaller than you might calculate if you made the assumption that it deformed and remained a circular cylinder.