1. Identify salts $W$ and $X$

Fusion of $MnO_2$ with $KOH$ in presence of $O_2$ gives potassium manganate: $2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$ So,

• $W = K_2MnO_4$
• Manganese-containing ion in $W$ is $Y = MnO_4^{2-}$ (manganate ion)

On electrolytic oxidation of alkaline manganate solution, it forms permanganate: $MnO_4^{2-} \xrightarrow[\text{electrolytic oxidation}]{} MnO_4^-$ Hence,

• $X =$ permanganate salt (such as $KMnO_4$)
• Manganese-containing ion in $X$ is $Z = MnO_4^-$

2. Examine geometry and colour of $Y$ and $Z$

The ions are:

• $Y = MnO_4^{2-}$
• $Z = MnO_4^-$

Both are oxoanions of manganese with four oxygens around Mn, so both have tetrahedral geometry.

Both are coloured:

• $MnO_4^{2-}$ is green
• $MnO_4^-$ is purple

So Option A is correct.

3. Check magnetic nature of $Y$ and $Z$

For $Y = MnO_4^{2-}$

Let oxidation state of Mn be $x$: $x + 4(-2) = -2 \implies x = +6$ Mn: $[Ar]3d^54s^2$ So $Mn^{6+}$ is: $3d^1$ Thus it has one unpaired electron  paramagnetic.

For $Z = MnO_4^-$

$x + 4(-2) = -1 \implies x = +7$ So $Mn^{7+}$ is: $3d^0$ Thus it has no unpaired electrons  diamagnetic.

Option B says: "Y is diamagnetic while Z is paramagnetic" which is exactly opposite.

So Option B is incorrect.

4. Check possibility of $\pi$-bonding

In both $MnO_4^{2-}$ and $MnO_4^-$, bonding is described with $p\pi-d\pi$ interaction between filled $p$-orbitals of oxygen and vacant/available $d$-orbitals of manganese.

Hence Option C is correct.

5. Check disproportionation of $Y$ in acidic medium

$Y = MnO_4^{2-}$, i.e. manganate ion. It is stable in alkaline medium but in acidic/neutral medium it undergoes disproportionation: $3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$ Thus manganate gives:

• permanganate, $MnO_4^-$ i.e. $Z$
• $MnO_2$

So Option D is correct.

6. Final selection

Correct options are: $\boxed{A,\ C,\ D}$

This matches the stored correct answer.