What is the oxidation state (number) of Phosphorous (P) in PO43- ? The oxidation number of O in this ion is -2.?

As the ion does not contain any element which has high priority in fixed oxidation number (i.e. Group I metals, Group II metals, aluminum, fluorine or hydrogen), the oxidation number of O is -2.

PO₄³⁻ is polyatomic ion, and thus
Total oxidation numbers = charge on the ion
(Oxidation number of P) + (Oxidation number of O) × 4 = -3
(Oxidation number of P) + (-2) × 4 = -3
(Oxidation number of P) - 8 = -3
Oxidation number of P = +5

The answer: +5