Step 1: Identify the Polynomial and Its Zeroes.
We are given $p(x) = 2x^2 - 7x + 6$ with zeroes $\alpha$ and $\beta$. We need to find $\dfrac{1}{\alpha} + \dfrac{1}{\beta}$.
Step 2: Apply Vieta's Formulas.
For a quadratic $ax^2 + bx + c = 0$, the sum and product of roots are: \[ \alpha + \beta = -\frac{b}{a}, \quad \alpha \cdot \beta = \frac{c}{a} \] Here $a = 2$, $b = -7$, $c = 6$, so: \[ \alpha + \beta = -\frac{-7}{2} = \frac{7}{2} \] \[ \alpha \cdot \beta = \frac{6}{2} = 3 \]
Step 3: Rewrite the Required Expression.
Combine the fractions: \[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha \cdot \beta} = \frac{\alpha + \beta}{\alpha \beta} \]
Step 4: Substitute the Values.
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\dfrac{7}{2}}{3} = \frac{7}{2} \times \frac{1}{3} = \frac{7}{6} \]
Step 5: Verify Using Root Values.
The roots of $2x^2 - 7x + 6 = 0$ can be found: $x = \dfrac{7 \pm \sqrt{49 - 48}}{4} = \dfrac{7 \pm 1}{4}$, giving $x = 2$ or $x = \dfrac{3}{2}$. So $\dfrac{1}{2} + \dfrac{2}{3} = \dfrac{3}{6} + \dfrac{4}{6} = \dfrac{7}{6}$. This confirms our answer.
Step 6: Match with Options.
$\dfrac{7}{6}$ corresponds to option (4).
\[ \boxed{\dfrac{7}{6}} \]