Explanation

(c)
In $\mathrm{\Delta}A\ BC,\ \ \overrightarrow{\mathbf{AB}} + \overrightarrow{\mathbf{BC}} + \overrightarrow{\mathbf{AC}}$
$\mathbf{\Longrightarrow \ }\overrightarrow{\mathbf{AC}} = \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}$
$AD$ is parallel to BC and AD $= 2\ BC$
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$\therefore\ \overrightarrow{\mathbf{AD}} = 2\overrightarrow{\mathbf{b}}$
In $\mathrm{\Delta}ACD,\ \ \overrightarrow{\mathbf{AC}} + \overrightarrow{\mathbf{CD}} = \overrightarrow{\mathbf{AD}}$
$\Longrightarrow \ \overrightarrow{\mathbf{CD}} = 2\overrightarrow{\mathbf{b}} - \left( \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} \right) = \overrightarrow{\mathbf{b}} - \overrightarrow{\mathbf{a}}$
Now, $\overrightarrow{\mathbf{CE}} = \overrightarrow{\mathbf{CD}} + \overrightarrow{\mathbf{DE}} = \overrightarrow{\mathbf{b}} - 2\overrightarrow{\mathbf{a}}$