Put $b=a^{1/4}$
$y=x^{1/4}$
The given limit can be written as
$\lim\limits_{y=a^{1/4}}\bigg(\bigg[\big(\large\frac{b^2+y^2}{b-y}\big)^{-1}-\frac{2by}{y^3-by^2+b^2y-b^3}\bigg]^{-1}-$$2^{\large 1/2\log_4 b^4}\bigg)^8$
$\Rightarrow \lim\limits_{y\to a^{1/4}}\bigg(\bigg[\large\frac{b-y}{b^2+y^2}-\frac{2by}{(y-b)(b^2+y^2}\bigg]^{-1}-$$b\bigg)^8$
$\Rightarrow \lim\limits_{\large y\to a^{1/4}}\bigg(\big[\large\frac{1}{b-y}\big]^{-1}$$-b\bigg)^8$
$\Rightarrow \lim\limits_{\large y\to a^{1/4}}y^8=a^2$
Hence (c) is the correct answer.