Actually the answer is far from obvious. At the exact center of the earth, the gravitational attraction in any direction is canceled out (assuming that the Earth was purely spherical, rather than the oblate spheroid that it is) by the attraction from an equal attraction in the other direction. If the Earth was hollow save for a thin shell, this would be true at any point within the earth, as it can be shown via calculus (I'll leave that exercise to you, but you can find it in any elementary physics book), that so long as you are within the sphere, there will be no force acting on you. However, anything outside of the shell see gravity as a point source at the center of that shell.
However if the Earth is filled (as it is), the moment that you move away from that center point, things get interesting. Consider the sphere r meters in radius centered on the Earth's center of gravity. This sphere now has a mass of m_{s} = (4/3)πr^{3}ρ, where ρ is the density of the sphere, which means the gravitational attractions becomes f = Gm_{s}m_{p}/r^{2} = (4/3)Gπr^{3}m_{p}ρ/r^{2} = 4Gπrm_{p}ρ/3.
Now the mass in question (m_{p}) is not the mass of the person, but the mass of the column of magma and rock. To a first approximation, for an area A, the mass will be m_{p} = A(r_{g} - r)ρ, where r_{g} is the radius of the earth. Substitution, you get, f = 4GπrA(r_{g} - r)ρ^{2}/3, or the pressure P = f/A = 4Gπr(r_{g} - r)ρ^{2}/3.
Now there are admittedly a number of simplifications here. ρ for instance, is technically a function ρ(r) which varies depending upon the properties of the material and the pressure acting on that property, but that's probably getting more complex than its worth. What it does imply is that the greatest pressure is not, in fact, at the center, but at the midpoint between the surface and the center of the earth.