Explanation

(c)
Let angle between $\overrightarrow{\mathbf{a}}\$and $\overrightarrow{\mathbf{b}}$ be $\theta_{1}.$ $\overrightarrow{\mathbf{c}}$ and $\overrightarrow{\mathbf{d}}$ be $\theta_{2}$ and $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{d}}$ beθ
Since, $\left( \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right) \bullet \left( \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{d}} \right) = 1$
$\Longrightarrow \sin{\theta_{1} \bullet \sin{\theta_{2} \bullet \cos{\theta = 1\ \ \ \left( \because\left| \overrightarrow{\mathbf{a}} \right| = \left| \overrightarrow{\mathbf{b}} \right| = \left| \overrightarrow{\mathbf{c}} \right| = \left| \overrightarrow{\mathbf{d}} \right| = 1 \right)}}}$
$\Longrightarrow \ \theta_{1} = 90{^\circ} \bullet \theta_{2} = 90{^\circ},\ \theta = 0{^\circ}$
$\Longrightarrow \overrightarrow{\mathbf{a}}\ \bot\overrightarrow{\mathbf{b}},\overrightarrow{\mathbf{c}}\ \bot\overrightarrow{\mathbf{d}},(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})||(\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{d}})$
So, $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} = k\left( \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{d}} \right)$ and $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} = k\left( \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{d}} \right)$
$\Longrightarrow \left( \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right) \bullet \overrightarrow{\mathbf{c}} = k\left( \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{d}} \right) \bullet \overrightarrow{\mathbf{c}}$
and $\left( \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right) \bullet \overrightarrow{\mathbf{d}} = k\left( \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{d}} \right) \bullet \overrightarrow{\mathbf{d}}$
$\Longrightarrow \lbrack\overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{c}}\rbrack = 0\ and\ \lbrack\overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{d}}\rbrack = 0$
$\Longrightarrow \ \overrightarrow{\mathbf{a}},\overrightarrow{\mathbf{b}},\overrightarrow{\mathbf{c}}$ and $\overrightarrow{\mathbf{a}},\overrightarrow{\mathbf{b}},\overrightarrow{\mathbf{d}}$ are coplanar vector so option (A) and (B) are incorrect.
Let $\overrightarrow{\mathbf{b}}||\overrightarrow{\mathbf{d}} \Longrightarrow \overrightarrow{\mathbf{b}} = \pm \overrightarrow{\mathbf{d}}$
As $\left( \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right) \bullet \left( \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{d}} \right) = 1 \Longrightarrow \left( \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right) \bullet \left( \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{b}} \right) = \pm 1$
$\Longrightarrow \left\lbrack \overrightarrow{\mathbf{a}}\ \times \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{c}}\ \overrightarrow{\mathbf{b}} \right\rbrack = \pm 1$
$\Longrightarrow \left\lbrack \overrightarrow{\mathbf{c}}\ \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right\rbrack = \pm 1$
$\Longrightarrow \ \ \overrightarrow{\mathbf{c}} \bullet \left\lbrack \overrightarrow{\mathbf{b}} \times \left( \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right) \right\rbrack = \pm 1$
$\Longrightarrow \ \ \overrightarrow{\mathbf{c}} \bullet \left\lbrack \overrightarrow{\mathbf{a}} - \left( \overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{a}} \right)\overrightarrow{\mathbf{b}} \right\rbrack = \pm 1$
$\Longrightarrow \ \overrightarrow{\mathbf{c}}\$∙ $\overrightarrow{\mathbf{a}} = \pm 1$ $\left( \because\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}} = 0 \right)$
Which is a contradiction so option (c) is correct.
Let option (d) is correct
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$\Longrightarrow \ \overrightarrow{\mathbf{d}} = \pm \overrightarrow{\mathbf{a}}\ and\ \overrightarrow{\mathbf{c}} = \pm \overrightarrow{\mathbf{b}}$
$As\ \ \ \ \left( \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right) \bullet \left( \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{d}} \right) = 1$
$\Longrightarrow \ \left( \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right) \bullet \left( \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}} \right) = \pm 1$
Which is a contradiction so option (d) is incorrect.
Alternate Option (c) and (d) may be observed from given in figure.