Reorganization during ionisation for d block elements
As your textbook notes, the relative energy of atomic orbitals is not fixed as you start ionizing your atom. As you ionize, all orbitals drop in energy. However, higher angular momentum orbitals drop more than lower ones, and specifically in your case, the $\mathrm{3d}$ orbitals drop below the $\mathrm{4s}$ orbitals. So consider vanadium, the atom has a $\mathrm{[Ar]4s^23d^3}$ configuration, but after you ionize it, the $\mathrm{3d}$ orbitals drop below the $\mathrm{4s}$ ones and you get $\mathrm{[Ar]3d^4}$ as the electron configuration of the $\ce{V+}$ ion. So your measured ionization energy includes both the energy it takes to remove an electron from $\ce{V}$ and also some reorganization energy. You know there is some reorganization energy since two electrons (both $\mathrm{4s}$ ones) have "moved" but you only removed one.

Now, what about $\ce{Cr}$? Recall that $\ce{Cr}$ is a bit special and has a $\mathrm{[Ar]3d^54s^1}$ configuration when neutral. After we remove an electron (in this case, a $\mathrm{4s}$ one), we have an $\mathrm{[Ar]3d^5}$ configuration for $\ce{Cr+}$. Note that no electrons had to move around besides the one we removed. So there is no reorganization energy in the case of $\ce{Cr}$, as your textbook notes.