Assume \(13! = x\)

Now, \(a = \frac{x^{16} - x^{8}}{x^{8} + x^{4}}\)

Taking \(x^{8}\) common from the Numerator and \(x^{4}\) from the Denominator:

\(a = \frac{x^{8}(x^{8} - x^{1})}{x^{4}(x^{4} + 1)}\)

Simplify:

\(a = \frac{x^{4}(x^{8} - x^{1})}{(x^{4} + 1)}\)

Break \((x^{8} - x^{1})\) into \((x^{4} + 1)(x^{4} - 1)\)

\(a = \frac{x^{4}(x^{4} + 1)(x^{4} - 1)}{(x^{4} + 1)}\)

\(a = x^{4}(x^{4} - 1)\)

So, \(\frac{a}{x^{4}} = \frac{x^{4}(x^{4} - 1)}{x^{4}} = x^{4} - 1\)

NOTE: \(13! = (13)(12)(11)(10)(9) ..... (1)\), has units digit as \(0\) only

Which means, \((13!)^{4} - 1 = (---------0) - 1 = ----------9\)

Hence, option E