Step 1: Combine the sum and product relations into a single ratio, avoiding solving for $\alpha$ directly.
Let the zeroes be $\alpha$ and $6\alpha$. By Vieta's formulas for $px^2 - 14x + 8$:
\[ \text{Sum} = \alpha + 6\alpha = 7\alpha = \frac{14}{p}, \qquad \text{Product} = \alpha \times 6\alpha = 6\alpha^2 = \frac{8}{p} \]
Step 2: Square the sum and divide by the product, so that $\alpha$ cancels out completely.
\[ \frac{(\text{Sum})^2}{\text{Product}} = \frac{(7\alpha)^2}{6\alpha^2} = \frac{49\alpha^2}{6\alpha^2} = \frac{49}{6} \]
The left side, written using $p$, is:
\[ \frac{\left(\frac{14}{p}\right)^2}{\frac{8}{p}} = \frac{\frac{196}{p^2}}{\frac{8}{p}} = \frac{196}{p^2} \times \frac{p}{8} = \frac{196}{8p} \]
Step 3: Set the two expressions equal.
\[ \frac{196}{8p} = \frac{49}{6} \]
Step 4: Cross multiply and solve for $p$.
\[ 196 \times 6 = 49 \times 8p \]
\[ 1176 = 392p \]
\[ p = \frac{1176}{392} = 3 \]
Final Answer:
The value of $p$ is 3, which corresponds to option (A).
\[ \boxed{p = 3} \]