Answer
The equation that shows the process for the fourth ionization energy of zirconium is $$Zr^{3+}\to Zr^{4+}+e^-$$
Work Step by Step
We already saw that the first ionization is the process to remove the first electron from a neutral atom, and the second ionization is the process to remove the second electron from an already-lost-one-electron ion.
Therefore, the fourth ionization of zirconium $(Zr)$ would be the process to remove the fourth electron from a zirconium ion, which has already lost 3 electrons before.
That means the equation that shows the process for the fourth ionization energy of zirconium is $$Zr^{3+}\to Zr^{4+}+e^-$$
