Explanation

(c)
If there is four $\sigma -$ bonds, hybridisation is $sp^{3},$ if three $\sigma -$ bonds, hybridisation is $sp^{2}$ and if two $\sigma -$ bonds, hybridisation is sp.
(a) $CH_{2} = C = CH_{2}\ \$
$sp^{2}$ sp $sp^{2}$
(b) $CH_{3} - CH = CH - {CH}_{2}^{+}$
$sp^{3}\ \ \ \ \ \ sp^{2}\ \ \ \ \ \ sp^{2}\ \ \ \ \ \ sp^{2}$
(c) $CH_{3} - C \equiv C - {CH}_{2}^{+}$
$sp^{3}\ \ \ \ \ \ sp\ \ \ \ \ \ sp\ \ \ \ \ \ sp^{2}$
(d) $CH_{3} - CH = CH - {CH}_{2}^{-}$
$sp^{3}\ \ \ \ \ \ sp^{2}\ \ \ \ \ \ sp^{2}\ \ \ \ \ \ sp^{3}$
(e) $CH_{2} = CH - CH = CH_{2}$
$sp^{2}\ \ \ \ \ \ sp^{2}\ \ \ \ \ \ sp^{2}\ \ \ \ \ \ sp^{2}$
Hence, in $CH_{3} - C \equiv C - {CH}_{2}^{+},$ all the three types of hybrid carbons are present.