Explanation

(c)
Hybridisation=$\frac{1}{2}$ [ no. of electron in valence shell + no. of monovalent atoms-charge on cation+ charge on anion]
$H_{2}O$
$H$=$\frac{1}{2}$ (6+2+0-0)=$\frac{8}{2}$=4
$\therefore$ $sp^{3}$hybridisation
$CH_{4}$
$H$=$\frac{1}{2}$ (4+4+0-0)
$\ \ \ \$=$\frac{8}{2}$=4
$\therefore$ $sp^{3}$hybridisation
$BCl_{3}$
$H$=$\frac{1}{2}$ (3+3+0-0)
$\ \ \$=$\frac{6}{2}$=3
$\therefore$ $sp^{3}$hybridisation
$NH_{3}$
$H$=$\frac{1}{2}$ (5+3+0-0)
$\ \ \$=$\frac{8}{2}$=4
$\therefore$ $sp^{3}$hybridisation
$\therefore$ (c) is correct answer.