Equation of hyperbola $x^2-4 y^2-4=0$
$$
\frac{x^2}{4}-\frac{y^2}{1}=1
$$
Then, $a=2, b=1$
Parametric point on hyperbola $=(a \sec \theta, b \tan \theta)$
Then, $\mathrm{P}\left(\frac{\pi}{4}\right)=\left(2 \sec \left(\frac{\pi}{4}\right), 1 \tan \left(\frac{\pi}{4}\right)\right)$
$$
=(2 \sqrt{2}, 1)
$$
Similarly, Q $\left(\frac{5 \pi}{4}\right)=\left(2 \sec \left(\frac{5 \pi}{4}\right), 1 \tan \left(\frac{5 \pi}{4}\right)\right)$
$$
\begin{aligned}
& =(-2 \sqrt{2}, 1) \\
& \mathrm{R}\left(\frac{3 \pi}{4}\right)=(-2 \sqrt{2},-1) \\
& \mathrm{T}\left(\frac{7 \pi}{4}\right)=(2 \sqrt{2},-1)
\end{aligned}
$$
Area of quadrilateral PQRT(sq. unit)
$$
\begin{aligned}
& =\frac{1}{2}\left\{\left(x_1-x_3\right)\left(y_2-y_4\right)-\left(x_2-x_4\right)\left(y_1-y_3\right)\right\} \\
& =\frac{1}{2}\{(2 \sqrt{2}+2 \sqrt{2})(1+1)-(-2 \sqrt{2}-2 \sqrt{2})(1+1)\} \\
& =\frac{1}{2}\{4 \sqrt{2} \times 2+4 \sqrt{2} \times 2\} \\
& =\frac{1}{2} \times 16 \sqrt{2}=8 \sqrt{2}
\end{aligned}
$$
So, option (d) is correct.