The electron configuration for C's ground state is 1s22s22p1x2p1y2p0z.
In order for C to be able to form 4 bonds, one electron from the 2s orbital is promoted to the empty 2pz orbital - this is referred to as the promoted state; this leads to the formation of hybrid orbitals - one s orbital mixes with the three p orbitals to form four sp3 hybrid orbitals.
This is what happens to the C atom in methane. Once the hybrid orbitals are formed, you can no longer refer to any of the three p orbitals. Here's a diagram of the formation of the sp3 hybrid orbitals:
)
The four sp3 orbitals now form sigma bonds with four H atoms to produce the tetrahedral-shaped methane molecule.
)
So, to answer your question, the 2pz orbital is not drawn because the C atom only has sp3 orbitals in the CH4 molecule.
Good explanation, nice figures can be found here: http://chemwiki.ucdavis.edu/Organic_Chemistry/Fundamentals/Hybrid_Orbitals
