Relation of liquid pressure with breadth 
Firstly, I would like to clear out that, my question is not very clear. I myself am not sure what problems I have on understanding this. So, I'll try my best to explain where I have confusions.
My confusions lies with this formula,
 $$ P= h \rho g$$
More specifically, I am confused about pressure when very thin pipes lead into very large tanks. For example, in a tank full of water there is a small ball, this ball experiences a pressure, P, due to the water above it. As the walls just above the ball of the tank becomes narrower, P, shouldn't change, and just before the walls conjoin, there is just an array of molecules in the "tube", and the pressure is still P. Once the walls conjoins, pressure P disappears. For some reason this seems absurd to me. In an instant, all the pressure is gone. I feel like something is missing but I am not sure what.
I hope you have been able to understand my confusion, because I haven't. Please help.
 A: 
Once the walls conjoins, pressure P disappears.

Not true.
The important point is that your formula ($P = h \rho g$) shows the increase in pressure from the top of a fluid to the bottom of a fluid.  Also, a lot of time we don't necessarily mean absolute pressure, but the pressure excess over atmospheric.
When the tube is open, your formula gives you the increase in pressure over the height of the tube.  And since it's open, you know the pressure at the top of the tube is 1atm.  
When you close the tube, the top of your (shortened) water column is now the top of the box.  But since the box is closed, you don't directly know what the pressure at the top is.  The absolute pressure at the bottom of the box can still be the same because the box is continuing to squeeze the fluid at a pressure above 1atm.

if there was no atmospheric pressure to begin with, how would there be atmospheric pressure after the box closes?

While the tube is there, we can determine the pressure at the level of the bottom of the tube.  If the tube has height $h$, then the pressure at the top of the box/bottom of the tube is $h \rho g$.  
When you seal the box (or close a valve), this pressure is still present.  It's just that instead of it coming from the water in the tube, it's coming from the sides of the vessel due to the quantity of water inside.
As neither before nor after closing the vessel is the pressure due to the atmosphere, it would be wrong to call this "atmospheric pressure".  It's a pressure due to the vessel, and the amount of pressure was set by the height of water in the tube before closing the valve.
Think of a bike tire.  You push some air in, you close the valve.  The fluid inside the tire remains pressurized.
In your example you push some water in (because the height of the tube is pushing down on water inside), then you close the valve and the fluid inside remains pressurized.
A: Pressure in liquids is a macroscopic property related to the equilibrium distance between molecules. That distance depends on their potential and kinetic energy. Here it is assumed that any non liquid molecule is far away to make any meaningful contribution.
If a pipe has a cross section small enough, the potential energy is no more only a function of the liquid molecules. The liquid-solid potential component must be taken in account, and its contribution increases gradually while the pipe diameter shrinks.
