**Step 1 — Let the zeroes be $\alpha$ and $6\alpha$.**

**Step 2 — Write the sum of the zeroes.**
$$\alpha + 6\alpha = -\frac{(-14)}{p} \ \Rightarrow\ 7\alpha = \frac{14}{p} \ \Rightarrow\ \alpha = \frac{2}{p} \qquad \dots(i)$$

**Step 3 — Write the product of the zeroes.**
$$\alpha \times 6\alpha = \frac{8}{p} \ \Rightarrow\ 6\alpha^2 = \frac{8}{p} \qquad \dots(ii)$$

**Step 4 — Substitute (i) into (ii).**
$$6\left(\frac{2}{p}\right)^2 = \frac{8}{p} \ \Rightarrow\ \frac{24}{p^2} = \frac{8}{p}$$

**Step 5 — Solve for $p$.**
$$24p = 8p^2 \ \Rightarrow\ 8p^2 - 24p = 0 \ \Rightarrow\ 8p(p - 3) = 0$$

Since $p \neq 0$ (otherwise it is not a quadratic polynomial),
$$p = 3$$

**Verification :** For $3x^2 - 14x + 8$, zeroes are $\dfrac{2}{3}$ and $4$, and $4 = 6 \times \dfrac{2}{3}$ ✔

$$\boxed{p = 3}$$