Let P = \ θ ∈ [0, 4π] : ^2θ ≠ 1\ and S = \ a ∈ ℤ : 2( ^8θ - ^8θ) 2θ...

Given the expression $2(\cos^8\theta - \sin^8\theta)\sec 2\theta = a^2$.

We can simplify the term $\cos^8\theta - \sin^8\theta$ as follows:
$\cos^8\theta - \sin^8\theta = (\cos^4\theta - \sin^4\theta)(\cos^4\theta + \sin^4\theta)$
$\Rightarrow (\cos^2\theta - \sin^2\theta)(\cos^2\theta + \sin^2\theta)((\cos^2\theta + \sin^2\theta)^2 - 2\sin^2\theta\cos^2\theta)$
$\Rightarrow \cos 2\theta \times 1 \times \left(1 - \dfrac{1}{2}\sin^2 2\theta\right)$

Substitute this back into the given equation:
$2 \left[ \cos 2\theta \left(1 - \dfrac{1}{2}\sin^2 2\theta\right) \right] \dfrac{1}{\cos 2\theta} = a^2$

Since $\theta \in P$, we have $\tan^2\theta \neq 1$, which implies $\cos 2\theta \neq 0$. Thus, we can safely cancel $\cos 2\theta$:
$2 \left(1 - \dfrac{1}{2}\sin^2 2\theta\right) = a^2$
$\Rightarrow 2 - \sin^2 2\theta = a^2$

We know that for any real $\theta$, $0 \le \sin^2 2\theta \le 1$.
However, since $\tan^2\theta \neq 1$, $\sin^2 2\theta \neq 1$.
Therefore, $0 \le \sin^2 2\theta < 1$.

Multiplying by $-1$ and adding $2$, we get the range of $a^2$:
$1 < 2 - \sin^2 2\theta \le 2$
$\Rightarrow 1 < a^2 \le 2$

Since $a \in \mathbb{Z}$, $a^2$ must be a perfect square integer. The only integer in the interval $(1, 2]$ is $2$, so $a^2 = 2$.

But there is no integer $a$ such that $a^2 = 2$.

Thus, the set $S$ contains no elements, meaning $S = \emptyset$.

Therefore, $n(S) = 0$.

Answer: $0$