Hybridisation of the underline atom changes in:  

$$\begin{gathered} \text{ Hybridisation }=\frac{1}{2}\left[\left(\begin{array}{l}\text{ No. of electrons }\\ \text{ in valence }\\ \text{ shell of atom }\end{array}\right)+\right. \\ \left(\begin{array}{l}\text{ No.of monovalent }\\ \text{ atoms around it }\end{array}\right)-\left(\begin{array}{l}\text{ charge on }\\ \text{ cation }\end{array}\right)+ \\ \left(\begin{array}{l}\text{ charge on }\\ \text{ anion }\end{array}\right) \end{gathered}$$

$$\begin{gathered} \text{ (a) }\mathrm{For}{\mathrm{AlH}}_3\text{, } \\ \text{ Hybridisation of }\mathrm{Al}\text{ atom }=\frac{1}{2}[3+3-0+0] \\ =3={\mathrm{sp}}^2 \\ \text{ For }{\mathrm{AlH}}_4^{-}\text{, } \\ \text{ Hybridisation of }\mathrm{Al}\text{ atom }=\frac{1}{2}[3+4-0+1] \\ =4={\mathrm{sp}}^3 \end{gathered}$$
$$\begin{gathered} \text{ (b) }{\mathrm{ForH}}_2\mathrm{O}\text{, } \\ \text{ Hybridisation of }\mathrm{O}\text{ atom } \\ =\frac{1}{2}[6+2-0+0]=4={\mathrm{sp}}^3 \\ \text{ For }{\mathrm{H}}_3{\mathrm{O}}^{+}\text{, Hybridisation of }\mathrm{O}\text{ atom } \\ =\frac{1}{2}[6+3-1+0]=4={\mathrm{sp}}^3 \end{gathered}$$
$$\begin{gathered} \text{ (c) For }{\mathrm{NH}}_3 \\ \text{ Hybridisation of }\mathrm{N}\text{ atom } \\ =\frac{1}{2}[5+3-0+0]=4={\mathrm{sp}}^3 \\ \text{ For }{\mathrm{NH}}_4^{+},\text{Hybridisation of }\mathrm{N}\text{ atom } \\ =\frac{1}{2}[5+4-1+0]=4={\mathrm{sp}}^3 \\ \text{ Thus hybridisation changes only in option (a). } \end{gathered}$$