Pressure versus mass and gravity I am not a physicist so I hope I can state my question in a coherent way.  I am trying to understand how pressure relates to gravity and mass.  For instance, pressure increases as we descend deeper into the Earth due to the increase of mass around us as we descend.  However, we also understand that gravity decreases as we descend because the the mass is no longer concentrated below our feet but distributed more evenly around us. So, what is creating pressure if gravity decreases with depth? I am sure I am missing some fundamental relationship between pressure, mass and gravity.  I hope my question is clear.
 A: You can imagine that the pressure at any point is equal to the sum of the pressure contribution of all "layers" above it, each of which are contributing due to the mass of and the gravity on that particular layer.
Because the earth does not have uniform density throughout (it has much of its mass concentrated at the core), gravitational acceleration does not decrease as you descend unless you get very, very low.  See Gravity beneath sea level
But even if it did decrease rapidly with depth, that effect would only decrease the pressure contribution from those lower layers.  The pressure contribution from the higher layers with the greater gravitational acceleration would remain.  Pressure would continue to increase with depth, just at a slower rate.

Now, there is zero gravitation at the very center but pressure remains from other layers above. Correct? What accounts for the pressure at the very center if gravity is zero? 

The same as above.  The pressure continues to grow as long as gravity is pointing in the same direction.  When you reach the center, $g=0$ so pressure no longer increases with distance.  But it doesn't magically go to zero.  
Remember pressure is force (over an area) and $F=ma$.  At any depth you can imagine a small volume of mass.  That volume is not accelerating, so the forces (pressures) must sum to zero.  That means any pressure from above plus the weight of the cube must be countered by pressure from below.
At the center, the weight of the volume is zero, but the pressure from any direction must be matched by an identical pressure from the opposite direction.
A: Take a ball of an ideal material which is incompressible and does not develop shear resistance: a perfect liquid.
Consider a cylinder with a small (infinitesimal) area, centrally drilled across the ball, till a distance $r$ from the center.
If the density is constant, the gravity will decrease moving towards the center, however the cylinder will have a weight $dW$ related to its $dA$, and such that  it gives a finite $p=dW/dA$. That's the pressure exerted at distance $r$ from the center, because without lateral forces the cylinder is sustained only on the base.
When $r=0$ the gravity will be $0$, but  a finite pressure will remain.
