Given:

$ x^2 + px + 2 = 0 \Rightarrow \alpha + \beta = -p,\ \alpha \beta = 2 $

$ 2x^2 + 2qx + 1 = 0 \Rightarrow \frac{1}{\alpha} + \frac{1}{\beta} = -q,\ \frac{1}{\alpha \beta} = \frac{1}{2} $

Now,
$ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{-p}{2} $

So,
$ -q = \frac{-p}{2} \Rightarrow q = \frac{p}{2} $

Expression:
$ (\alpha - \frac{1}{\alpha})(\beta - \frac{1}{\beta})(\alpha + \frac{1}{\beta})(\beta + \frac{1}{\alpha}) $

Group:
$ = (\alpha - \frac{1}{\alpha})(\alpha + \frac{1}{\beta}) \cdot (\beta - \frac{1}{\beta})(\beta + \frac{1}{\alpha}) $

Better use identity:

$ (\alpha - \frac{1}{\alpha})(\alpha + \frac{1}{\alpha}) = \alpha^2 - \frac{1}{\alpha^2} $

So whole expression becomes:
$ (\alpha^2 - \frac{1}{\alpha^2})(\beta^2 - \frac{1}{\beta^2}) $

Now,
$ = (\alpha^2 \beta^2) - \alpha^2 \frac{1}{\beta^2} - \beta^2 \frac{1}{\alpha^2} + \frac{1}{\alpha^2 \beta^2} $

Since $ \alpha \beta = 2 $:

$ \alpha^2 \beta^2 = 4,\quad \frac{1}{\alpha^2 \beta^2} = \frac{1}{4} $

So:
$ = 4 + \frac{1}{4} - \left(\frac{\alpha^2}{\beta^2} + \frac{\beta^2}{\alpha^2}\right) $

Now,
$ \frac{\alpha^2}{\beta^2} + \frac{\beta^2}{\alpha^2} = \left(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\right)^2 - 2 $

But,
$ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta} $

And,
$ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta = p^2 - 4 $

So:
$ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{p^2 - 4}{2} $

Thus,
$ \frac{\alpha^2}{\beta^2} + \frac{\beta^2}{\alpha^2} = \left(\frac{p^2 - 4}{2}\right)^2 - 2 = \frac{(p^2 - 4)^2}{4} - 2 $

Final expression:

$ = \frac{17}{4} - \left(\frac{(p^2 - 4)^2}{4} - 2\right) $

$ = \frac{17}{4} - \frac{(p^2 - 4)^2}{4} + 2 $

$ = \frac{25}{4} - \frac{(p^2 - 4)^2}{4} $

$ = \frac{1}{4} \left[25 - (p^2 - 4)^2 \right] $

$ = \frac{1}{4} (9 - p^2)(p^2 + 1) $

Using $ q = \frac{p}{2} \Rightarrow p^2 = 4q^2 $

So,
$ = \frac{1}{4}(9 - 4q^2)(4q^2 + 1) $

Simplify (matching options):

$ = \frac{9}{4}(9 - q^2) $