Explanation

(b)
(a) $NH_{3}\ \ \ \ + {\ \ \ H}^{+}\ \ \longrightarrow {NH}_{4}^{+}$
(3bp + 1lp) $(4bp \Longrightarrow sp^{3}\$hybridisation$)$
$\Longrightarrow sp^{3}\$hybridisation)
(b) $BF_{3} + {\ \ F}^{-} \longrightarrow {BF}_{4}^{-}$
(3bp + $sp^{2}\$hybridisation)
$(4bp \Longrightarrow sp^{3}\$hybridisation$)$
(c) $H_{2}O + {\ H}^{+} \longrightarrow H_{3}O^{+}$
(2bp + 2lp (3bp + 1lp
$\Longrightarrow sp^{3}\$hybridisation)$\ \ \ \Longrightarrow sp^{3}\$hybridisation)
(d) $CH \equiv CH + 2H_{2}\ \longrightarrow CH_{3} - CH_{3}$
sp sp $sp^{3}\ \ \ \ \ \ sp^{3}$
Hence, reaction given in option (b) involves the change of hybridisation from $sp^{2\ }$to$\ sp^{3}$.