If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, prove the following.

You correctly computed $$ (\alpha^2-1)(\beta^2-1)(\gamma^2-1)(\delta^2-1)=(1+q+s)^2-(p+r)^2 $$ only that this is not what was asked for.

But you were close: $\alpha^2+1 = (\alpha-i)(\alpha+i)$ etc, and therefore $$ (1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\delta^2) = P(i) P(-i) $$ which expands to $$ (1-ip-q+ir+s)(1+ip-q-ir+s) = (1-q+s)^2+(p-r)^2 \, . $$

Denoting $\alpha=x_1$, $\beta=x_2$, $\gamma=x_3$ and $\delta=x_4$, we have $$\small\prod_{i=1}^4(1+x_i^2)-\prod_{i=1}^4(1-x_i^2)=\small2(x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4)+2(x_1+x_2+x_3+x_4)$$ so taking out a factor of $x_1x_2x_3x_4$ from the first term in brackets, $$\prod_{i=1}^4(1+x_i^2)=2\left(\prod_{i=1}^4x_i\right)\left(\sum_{i=1}^4\frac1{x_i}\right)+2\sum_{i=1}^4x_i+\prod_{i=1}^4(1-x_i^2)$$ which is evaluable by Vieta's formulas. For the sum of reciprocals, see this post.

Another way using Transformation of equation:

We have $$x^4+qx^2+s=-x(px^2+r)$$

Squaring both sides

$$(x^2)^4+q^2(x^2)+s^2+2q(x^2)^3+2qs(x^2)+2s(x^2)^2=x^2(p^2(x^2)^2+r^2+2pr(x^2))$$

Replacing $x^2+1=y$

$$(y-1)^4+q^2(y-1)^2+s^2+2q(y-1)^3+2qs(y-1)+2s(y-1)^2=(y-1)(p^2(y-1)^2+r^2+2pr(y-1))$$

$$\iff y^4+\cdots+1+q^2+s^2-2q-2qs+2s+p^2+r^2-2pr=0$$

Using Vieta's formula $$\prod_{j=1}^4y_j=\dfrac{1+q^2+s^2-2q-2qs+2s+p^2+r^2-2pr}1=(1-q+s)^2+(p-r)^2$$