The oxidation state of the metal centre is $$+VII$$.
Explanation:
But the charge on the permanganate ion is $$−1$$. You have its potassium salt, i.e permanganate is the counterion to $$K^+$$, and thus it is reasonable that we have a $$MnO^−_4$$ ion. In other words, the charge on the overall ion is $$−1$$; it might lie on the metal or more likely the oxygen, but it is there.
When permanganate is reduced, it goes down to $$Mn(+II)$$, a $$d^5$$ system that is almost colourless. This makes the redox couple very useful for titrations because chemical change can be observed macroscopically in real time by the dramatic colour change.
$$MnO^-_4+8H^++5e^-\rightarrow Mn^{2+}+4H_2O$$