$\large z = \dfrac {1}{ 2} \left( \sqrt{3}-i \right )$
$\large z^2 =\dfrac 14 \left ( 3 -1 -2\sqrt{3}\,i \right ) = \dfrac 12 \left (1 - \sqrt{3}\,i \right )$
$\large z^4 =\dfrac 14 \left (1 - 3 - 2 \sqrt{3}\,i \right ) =\dfrac 12 \left (-1-\sqrt{3}\,i \right )$
$\large z^8 = \dfrac 14 \left (1 - 3 + 2\sqrt{3}\,i \right ) = \dfrac 12 \left ( -1+\sqrt{3}\,i\right )$
$\large z^{16} = \dfrac 14 \left (1-3-2\sqrt{3}\,i \right ) = \dfrac 12 \left (-1-\sqrt{3}\,i \right ) = \color{red}{z^4}$
$\large z^{32} =z^{16} \times z^{16} = z^4 \times z^4 = z^8$
$\large z^{64} = z^{32} \times z^{32} =z^8 \times z^8 = z^{16} = z^4$
$\large z^{95} =z^{64} \times z^{16} \times z^{15}$
$\large = z^4 \times z^4 \times z^{15}$
$\large =z^{16} \times z^7$
$\large =z^4 \times z^7$
$\large =z^8 \times z^2 \times z$
$\large =\dfrac 12 \left ( -1+\sqrt{3}\,i\right ) \times \dfrac 12 \left (1 - \sqrt{3}\,i \right ) \times \dfrac 12 \left ( \sqrt{3}-i \right )$
$\large =\dfrac 12 \left (\sqrt{3} + i \right )$
$\large i^{67} =i^{64} \times i^3$
$\large = 1 \times (-i)$
$\large =-i$
$\large z^{95} + i^{67} =\dfrac 12 \left (\sqrt{3} + i \right ) - i$
$\large =\dfrac 12 \left (\sqrt{3} - i \right )$
$\large =z$
$\large \left (z^{95} + i^{67} \right )^{97} = z^{97} = z^{95} \times z^{2}$
$\large =\dfrac 12 \left (\sqrt{3} + i \right ) \times \dfrac 12 \left (1 - \sqrt{3}\,i \right )$
$\large = \dfrac 12 \left ( \sqrt{3}-i \right )$
$\large = z$
Hence, option a is the correct answer.