The electron configuration for #C#'s ground state is #1s^(2)2s^(2)2p_x^(1)2p_y^(1)2p_z^(0)#.
In order for C to be able to form 4 bonds, one electron from the #2s# orbital is promoted to the empty #2p_z# 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 #sp^3# 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 #sp^3# hybrid orbitals:
The four #sp^3# orbitals now form sigma bonds with four #H# atoms to produce the tetrahedral-shaped methane molecule.
So, to answer your question, the #2p_z# orbital is not drawn because the #C# atom only has #sp^3# orbitals in the #CH_4# molecule.
Good explanation, nice figures can be found here: http://chemwiki.ucdavis.edu/Organic_Chemistry/Fundamentals/Hybrid_Orbitals
