Unequal ionization energies of methane
In the photoelectron spectroscopy (PES) experiment a sample is irradiated with photons of differing energy and the kinetic energy of any ejected electrons is measured.  By knowing the energy of the incoming photon and the kinetic energy of any ejected electrons, we can calculate the energy required to ionize the electron as the difference between these two recorded values. An electron will only be ejected if the photon energy is equal to or greater than the energy required to ionize an electron in the molecule.  See the section entitled "Photoelectron Spectroscopy" in this link if you would like to read more.

In the case of methane, all of the bonds are equivalent as required by symmetry.  There is only one $\ce{C-H}$ bond strength for methane.

When we run the PES experiment with methane we see 3 signals (as pictured in your link; two of the signals are due to photoionization of electrons from the $\ce{C-H}$ bond, and the third signal, at much higher photon energy, is due to photoionization of the carbon 1s electrons).  We know that the 4 bonds in methane can be described by 8 orbitals, 4 bonding, 4 antibonding.  We can transform these 8 localized orbitals into 8 molecular orbitals; the 4 bonding MO's are pictured in the link you provided.  Notice from the picture in your link that 3 of the 4 MO's are degenerate, but there is also a different 4th MO.  All of these bonding MO's are filled and ionization can occur from either one of the 2 types - hence 2 signals from the occupied $\ce{C-H}$ MO's (in the PES experiment, these are not sequential ionizations, e.g. these are not first and second ionizations; rather they are independent ionizations that will occur only when our incoming photon energy matches the energy of the orbital containing electrons).  During the PES experiment with methane, no bonds are broken, rather an electron from an orbital is ejected and the methyl radical cation is formed.