If a, b, c and d are the unit vectors such that (a × b) ·(c × d)=1...

Let angle between $\mathbf{a} and$\mathbf{b}$be$\theta_1, \mathbf{c} and $\mathbf{d}$ be $\theta_2$ and $\mathbf{a} \times \mathbf{b} and \mathbf{c} \times \mathbf{d}$ be $\theta$. Since, $(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=1\Rightarrow \quad \sin \theta_1 \cdot \sin \theta_2 \cdot \cos \theta=1\Rightarrow \theta_1=90^{{}^{\circ}}, \theta_2=90^{{}^{\circ}}, \theta=0^{{}^{\circ}}\Rightarrow \mathbf{a} \perp \mathbf{b}, \mathbf{c} \perp \mathbf{d},(\mathbf{a} \times \mathbf{b}) \|(\mathbf{c} \times \mathbf{d})$ So, $\mathbf{a} \times \mathbf{b}=k(\mathbf{c} \times \mathbf{d})$ and $\mathbf{a} \times \mathbf{b}=k(\mathbf{c} \times \mathbf{d})\Rightarrow \quad(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}=k(\mathbf{c} \times \mathbf{d}) \cdot \mathbf{c} and (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{d}=k(\mathbf{c} \times \mathbf{d}) \cdot \mathbf{d}\Rightarrow$\left[$$\begin{array}{lll}\mathbf{a} & \mathbf{b} & \mathbf{c}\end{array}$$\right]=0and \left[$\begin{array}{lll}\mathbf{a} & \mathbf{b} & \mathbf{d}\end{array}$\right]=0\Rightarrow \mathbf{a}, \mathbf{b}, \mathbf{c} and $\mathbf{a}, \mathbf{b}, \mathbf{d}$ are coplanar vectors, so options (a) and (b) are incorrect. Let $\quad \mathbf{b} \| \mathbf{d} \Rightarrow \mathbf{b}=\pm \mathbf{d}$ As $\quad(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=1(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{b})=\pm 1\Rightarrow \quad[\mathbf{a} \times \mathbf{b} \mathbf{c} \mathbf{b}]=\pm 1\Rightarrow \quad[\mathbf{c} \mathbf{b} \mathbf{a} \times \mathbf{b}]=\pm 1\Rightarrow \quad \mathbf{c} \cdot[\mathbf{b} \times(\mathbf{a} \times \mathbf{b})]=\pm 1\Rightarrow \quad \mathbf{c} \cdot[\mathbf{a}-(\mathbf{b} \cdot \mathbf{a}) \mathbf{b}]=\pm 1\Rightarrow \quad \mathbf{c} \cdot \mathbf{a}=\pm 1 \quad[\because \mathbf{a} \cdot \mathbf{b}=0]$ which is a contradiction, so option (c) is correct.
Let option (d) be correct. $$\begin{array}{ll} \Rightarrow& \mathbf{d}=\pm \mathbf{a} \text{\text{ and } } \mathbf{c}=\pm \mathbf{b}\\ \text{ As }& (\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=1 \\ \Rightarrow & (\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{b} \times \mathbf{a})=\pm1 \end{array}$$ which is a contradiction, so option (d) is incorrect. Alternatively option (c) and (d) may be observed from the given figure.