1. We need to check whether the hybridisation of the underlined atom changes in each conversion.

2. Option A: $\underline{Al}H_3 \to AlH_4^-$

• In $AlH_3$, aluminium forms 3 bond pairs and has no lone pair.
• Steric number $=3$, so geometry is trigonal planar.
• Hybridisation of $Al$ is $sp^2$.

Now for $AlH_4^-$:

• Aluminium is bonded to 4 hydrogens.
• Steric number $=4$, so geometry is tetrahedral.
• Hybridisation of $Al$ is $sp^3$.

So, in option A, hybridisation changes: $sp^2 \to sp^3$

Hence, A is correct.

3. Option B: $H_2\underline{O} \to H_3O^+$

For $H_2O$:

• Oxygen has 2 bond pairs and 2 lone pairs.
• Steric number $=4$.
• Hybridisation of $O$ is $sp^3$.

For $H_3O^+$:

• Oxygen has 3 bond pairs and 1 lone pair.
• Steric number $=4$.
• Hybridisation of $O$ is still $sp^3$.

So, in option B, hybridisation does not change.

Hence, B is incorrect.

4. Option C: $\underline{N}H_3 \to NH_4^+$

For $NH_3$:

• Nitrogen has 3 bond pairs and 1 lone pair.
• Steric number $=4$.
• Hybridisation of $N$ is $sp^3$.

For $NH_4^+$:

• Nitrogen has 4 bond pairs and 0 lone pairs.
• Steric number $=4$.
• Hybridisation of $N$ is still $sp^3$.

So, in option C, hybridisation does not change.

Hence, C is incorrect.

5. Option D: “in all cases”

Since only option A shows a change in hybridisation, option D is incorrect.

6. Final Answer

Only Option A is correct.

$\boxed{A}$