Binding energy vs. ionization energy
First some basics....
Lithium has three electrons.

$\ce{Li}$ electronic configuration: $1s^22s^1$
$\ce{Li -&gt; Li+ + e-\quad\quad\mathrm{Ionization\ Energy= 0.520\ MJ/mol}}$
$\ce{Li+}$ electronic configuration: $1s^2$
$\ce{Li+ -&gt; Li^2+ + e-\quad\quad\mathrm{Ionization\ Energy= 7.298\ MJ/mol}}$
$\ce{Li^2+}$ electronic configuration: $1s^1$
$\ce{Li^2+ -&gt; Li^3+ + e-\quad\quad\mathrm{Ionization\ Energy=11.815\ MJ/mol}}$
$\ce{Li^3+}$ electronic configuration: $1s^0$, i.e. no electrons
Ionization energies from Wikipedia

Thus the different ionization reactions remove the electrons from the outer most to the inner most electrons.
Below is an another image of the idealized spectra of lithium that has the binding energies stated. (from https://www.andrew.cmu.edu/course/09-105/S98Slides9.html)

$\mathrm{Binding\ Energy= 0.52\ MJ/mol}$ corresponds to $\ce{Li -&gt; Li+ + e-}$
or $1s^2 2s^1\ce{-&gt;} 1s^2$, hence the first ionization energy of a Li atom.
However $\mathrm{Binding\ Energy= 6.26\ MJ/mol}$ corresponds to $\ce{Li -&gt; Li+ + e-}$
or $1s^2 2s^1\ce{-&gt;} 1s^12s^1$, which is not the first or second ionization energy of a Li atom. Rather it is the binding energy of the 1s orbital in a neutral lithium atom.
Hence the photoelectron spectra is called the binding energy spectra because that is what it measures. Note that the sweep of the spectra would not have to cover the ionization energy. For example the spectra could have been taken from 5 MJ/mol to 100 MJ/mol.