Let $$\alpha, \beta$$ be the roots of the equation $$x^2 - x + p = 0$$ and $$\gamma, \delta$$ be the roots of the equation$$x^2 - 4x + q = 0$$, where $$p, q \in \mathbb{Z}$$. If $$\alpha, \beta, \gamma, \delta$$ are in G.P., then $$|p + q|$$  equals :

Let the four numbers be in the same order as stated: $$\alpha ,\beta ,\gamma ,\delta$$ form a geometric progression (G.P.) with common ratio $$r$$.

Hence
$$\beta = \alpha r,\qquad \gamma = \alpha r^{2},\qquad \delta = \alpha r^{3}$$

The roots of the first equation $$x^{2}-x+p=0$$ satisfy
$$\alpha + \beta = 1,\qquad \alpha \beta = p$$

The roots of the second equation $$x^{2}-4x+q=0$$ satisfy
$$\gamma + \delta = 4,\qquad \gamma \delta = q$$

Using the G.P. relations, write the two sums in terms of $$\alpha$$ and $$r$$:
$$\alpha + \beta = \alpha + \alpha r = \alpha(1+r) = 1 \qquad -(1)$$
$$\gamma + \delta = \alpha r^{2} + \alpha r^{3} = \alpha r^{2}(1+r) = 4 \qquad -(2)$$

Divide $$(2)$$ by $$(1)$$:
$$\frac{\alpha r^{2}(1+r)}{\alpha(1+r)} = r^{2} = 4 \;\Longrightarrow\; r = \pm 2$$

From $$(1)$$, $$\alpha = \dfrac{1}{1+2} = \dfrac{1}{3}$$.
Then $$p = \alpha^{2} r = \left(\dfrac{1}{3}\right)^{2}\!\cdot 2 = \dfrac{2}{9} \notin \mathbb{Z}$$. This violates the condition $$p \in \mathbb{Z}$$, so $$r = 2$$ is rejected.

From $$(1)$$, $$\alpha = \dfrac{1}{1-2} = -1$$.

Now compute the required products:
$$p = \alpha^{2} r = (-1)^{2}(-2) = -2$$
$$q = \gamma \delta = (\alpha r^{2})(\alpha r^{3}) = \alpha^{2} r^{5} = (-1)^{2}(-2)^{5} = -32$$

Both $$p$$ and $$q$$ are integers, so this case is admissible.

Finally,
$$p + q = -2 + (-32) = -34 \;\Longrightarrow\; |p+q| = 34$$

Therefore, the required value is $$34$$.

Option C which is: $$34$$