Let α and β be the roots of equation p x ^ 2 + qx + r = 0 , p ≠ 0 ....

$\dfrac{1}{\alpha} + \dfrac{1}{\beta} = 4$

$\dfrac{\alpha + \beta}{\alpha \beta} = 4$

$∴ \dfrac{- \dfrac{q}{p}}{\dfrac{r}{p}} = 4$

$∴ - \dfrac{q}{r} = 4$

$p, q, r$ are in A.P.

Let $p = a - d; q = a, r = a + d$.

$∴ \dfrac{- \left(a\right)}{a + d} = 4 ⇒ - a = 4a + 4d ⇒ d = \dfrac{- 5a}{4}$

$p = a + \dfrac{5a}{4} = \dfrac{9a}{4}; q = a, r = a - \dfrac{5a}{4} = - \dfrac{a}{4}$

$∴ \dfrac{9}{4}{x}^{2} + \dfrac{4}{4}x - \dfrac{1}{4} = 0$

$∴ 9 {x}^{2} + 4x - 1 = 0$

${\left|\alpha - \beta\right|}^{2} = {\left(\alpha + \beta\right)}^{2} - 4 \alpha \beta$

$= {\left(\dfrac{- 4}{9}\right)}^{2} - 4 \left(- \dfrac{1}{\text{9}}\right)$

$= \dfrac{1 6}{8 1} + \dfrac{4}{9} = \dfrac{5 2}{8 1}$

$∴ \left|\alpha - \beta\right| = \dfrac{2 \sqrt{1 3}}{9}$.

Hence, the correct answer is $\dfrac{2 \sqrt{1 3}}{9}$.