$$f(x) = {{\root 7 \of {p(729 + x)} - 3} \over {\root 3 \of {729 + qx} - 9}}$$

for continuity at $$x = 0$$, $$\mathop {\lim }\limits_{x \to 0} f(x) = f(0)$$

Now, $$\therefore$$ $$\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} {{\root 7 \of {p(729 + x)} - 3} \over {\root 3 \of {729 + qx} - 9}}$$

$$ \Rightarrow p = 3$$ (To make indeterminant form)

So, $$\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} {{{{\left( {{3^7} + 3x} \right)}^{{1 \over 7}}} - 3} \over {{{\left( {729 + qx} \right)}^{{1 \over 3}}} - 9}}$$

$$ = \mathop {\lim }\limits_{x \to 0} {{3\left[ {{{\left( {1 + {x \over {{3^6}}}} \right)}^{{1 \over 7}}} - 1} \right]} \over {9\left[ {{{\left( {1 + {q \over {729}}x} \right)}^{{1 \over 3}}} - 1} \right]}} = {1 \over 3}\,.\,{{{1 \over 7}\,.\,{1 \over {{3^6}}}} \over {{1 \over 3}\,.\,{q \over {729}}}}$$

$$\therefore$$ $$f(0) = {1 \over {7q}}$$

$$\therefore$$ Option (B) is correct.