[Win,odd Dhamnekar]

 Using Slater's rule Z_eff on  electrons in  3p orbital of Phosphorous is 4.8
http://calistry.org/calculate/slaterRuleCalculator




But If we use E_ionization=-E_n,I=[itex] \frac{Z_{eff}^2 \times R_H}{n^2} \Rightarrow Z_{eff}=\big( \frac{n^2 \times E_{n,I}}{R_H}\big)^{\frac12}, Z_{eff}[/itex] on electrons in 3p orbital of Phosphorous  are different.

e.g. 1st, 2nd, and 3rd [itex]E_{(3, I)}[/itex] of Phosphorous are 1011.8 kJ/mol, 1907 kJ/mol, 2914.1kJ/mol repectively. If we plug in these energy values in the  above formula, we get Z_eff=2.633, using 1st E_(3,I), Z_eff= 3.616 using 2nd E_(3,I) and Z_eff=4.47 using 3rd E_(3,I).

Why are there such discrepancies in computations of Z_eff using these two different methods?

[mjc123]

Where did you get the formula for ionization energy? It looks as if it applies to a Rydberg atom - an atom with a single valence electron outside a compact core, that behaves similarly to a hydrogen-like atom with charge Z_eff. It won't work for an atom with multiple valence electrons. The ionization energy is not equal to the energy level of the electron being removed, because when you remove that electron the energy levels of all the other electrons change.

[Win,odd Dhamnekar]

 I got the ionization formula in the following answer to the same(similar) question.

[mjc123]

If you define Z_eff in that way, it is quite different from Slater's definition, and it is not surprising that you get different values. In my view it is not a very meaningful definition, because as I said P is not a Rydberg atom. When you remove an electron the interactions between the remaining electrons change, so you get different Z_eff values for P⁺ and P²⁺. (You do with Slater as well.)