To determine which option corresponds to the scenario where the equivalent weight of $MnSO_4$ is half its molecular weight upon conversion, we need to analyze each oxidation process and find out the change in oxidation state of manganese (Mn) in each case. The equivalent weight of a compound in a given reaction depends on the change in oxidation number per atom of the element in question. It is calculated as the molecular weight divided by the number of electrons lost or gained in the reaction.

The molecular weight of $MnSO_4$ (Manganese(II) sulfate) is primarily irrelevant for the specifics of calculating equivalent weight in this manner; what's crucial is the stoichiometry of the electron transfer involved in its reaction.

Let's consider the oxidation state of Mn in $MnSO_4$, which is +2.

Option A: $Mn_2O_3$

Mn's oxidation state in $Mn_2O_3$ is +3. Thus, going from +2 in $MnSO_4$ to +3 in $Mn_2O_3$ involves a loss (or gain) of 1 electron per manganese atom. Given that there are two Mn atoms in $Mn_2O_3$, this would not match the stated requirement directly without considering the specifics of the individual reaction stoichiometry.

Option B: $MnO_2$

The oxidation state of Mn in $MnO_2$ is +4. The change from +2 to +4 implies a gain or loss of 2 electrons.

Option C: $MnO_4^-$ (Permanganate)

In $MnO_4^-$, the oxidation state of Mn is +7. The change from +2 in $MnSO_4$ to +7 in $MnO_4^-$ involves a change of 5 electrons.

Option D: $MnO_4^{2-}$

This oxidation state is not typically encountered for manganese and may represent a mistaken formulation, as Mn typically shows +7 oxidation state in permanganate $(MnO_4^-)$, not a doubly charged anion.

Given these considerations, the question's conditions mention that the equivalent weight of $MnSO_4$ becomes half of its molecular weight when it converts to one of these products. This would imply a situation where the number of electrons transferred per formula unit of $MnSO_4$ is equivalent to 2 (since the molecular weight divided by 2 equals the equivalent weight when each mole of $MnSO_4$ involves the transfer of 2 moles of electrons).

Thus, the correct answer is:

Option B: $MnO_2$, because the conversion from $Mn^{+2}$ to $Mn^{+4}$ involves a transfer of 2 electrons per manganese atom, which fits the condition that the equivalent weight is half the molecular weight in this context.