Let $$P = \{\theta \in [0, 4\pi] : \tan^2\theta \ne 1\}$$ and $$S = \{a \in \mathbb{Z} : 2(\cos^8\theta - \sin^8\theta)\sec 2\theta = a^2, \theta \in P\}$$. Then $$n(S)$$ is :

The given expression is
$$E = 2\bigl(\cos^{8}\theta-\sin^{8}\theta\bigr)\sec 2\theta.$$

Step 1 : Simplify $$\cos^{8}\theta-\sin^{8}\theta$$
Write it as a product of two squares:
$$\cos^{8}\theta-\sin^{8}\theta=(\cos^{4}\theta-\sin^{4}\theta)(\cos^{4}\theta+\sin^{4}\theta).$$
Now
$$\cos^{4}\theta-\sin^{4}\theta=(\cos^{2}\theta-\sin^{2}\theta)(\cos^{2}\theta+\sin^{2}\theta)=\cos 2\theta,$$
because $$\cos^{2}\theta+\sin^{2}\theta=1.$$
Hence
$$\cos^{8}\theta-\sin^{8}\theta=\cos 2\theta\bigl(\cos^{4}\theta+\sin^{4}\theta\bigr).$$

Step 2 : Insert this in $$E$$
$$E = 2\cos 2\theta\bigl(\cos^{4}\theta+\sin^{4}\theta\bigr)\sec 2\theta = 2\bigl(\cos^{4}\theta+\sin^{4}\theta\bigr)$$ because $$\cos 2\theta\sec 2\theta = 1.$$

Step 3 : Convert $$\cos^{4}\theta+\sin^{4}\theta$$ to a single-angle form
First use $$\sin^{2}2\theta=4\sin^{2}\theta\cos^{2}\theta.$$
$$$ \cos^{4}\theta+\sin^{4}\theta =( \cos^{2}\theta+\sin^{2}\theta )^{2}-2\sin^{2}\theta\cos^{2}\theta =1-\tfrac12\sin^{2}2\theta. $$$
Since $$\sin^{2}2\theta=\tfrac12(1-\cos4\theta),$$ we get
$$\cos^{4}\theta+\sin^{4}\theta =1-\tfrac12\!\bigl[\tfrac12(1-\cos4\theta)\bigr] =\tfrac34+\tfrac14\cos4\theta =\frac{3+\cos4\theta}{4}.$$

Step 4 : Final form of $$E$$
$$E = 2\bigl(\cos^{4}\theta+\sin^{4}\theta\bigr) = 2\left(\frac{3+\cos4\theta}{4}\right) =\frac{3+\cos4\theta}{2}.$$

Step 5 : Impose the condition $$E=a^{2}$$ with $$a\in\mathbb{Z}$$
The range of $$\cos4\theta$$ is $$[-1,1]$$, so
$$2 \le 3+\cos4\theta \le 4.$$
Therefore
$$1 \le a^{2}=\frac{3+\cos4\theta}{2}\le 2.$$
The only integer square lying in $$[1,2]$$ is $$1$$. Hence
$$a^{2}=1\quad\Longrightarrow\quad a=\pm1.$$

Step 6 : Check whether such $$a$$ actually occur with $$\theta\in P$$
For $$a^{2}=1$$ we need
$$\frac{3+\cos4\theta}{2}=1\;\Longrightarrow\;3+\cos4\theta=2 \;\Longrightarrow\;\cos4\theta=-1.$$
Thus
$$4\theta=(2k+1)\pi\;\Longrightarrow\;\theta=\frac{(2k+1)\pi}{4},\qquad k\in\mathbb{Z}.$$

Step 7 : Verify the restriction $$\tan^{2}\theta\ne1$$
For $$\theta=\dfrac{(2k+1)\pi}{4}$$ we have
$$\theta=\frac{\pi}{4}+k\frac{\pi}{2}\quad\Longrightarrow\quad \tan\theta=\pm1 \;\Longrightarrow\;\tan^{2}\theta=1.$$
These values of $$\theta$$ are excluded from the set $$P$$ (because $$\tan^{2}\theta\ne1$$ was required).
Hence no permissible $$\theta$$ satisfies $$E=a^{2}$$.

Conclusion
No integer $$a$$ arises from any $$\theta\in P$$, so the set $$S$$ is empty and
$$n(S)=0.$$
Option A.