Let $$a$$ and $$b$$ be two distinct positive real numbers. Let 11th term of a GP, whose first term is $$a$$ and third term is $$b$$, is equal to $$p^{th}$$ term of another GP, whose first term is $$a$$ and fifth term is $$b$$. Then $$p$$ is equal to

GP1: first term $$a$$, third term $$b$$. Common ratio $$r_1 = \sqrt{b/a}$$. 11th term = $$a \cdot r_1^{10} = a(b/a)^5 = b^5/a^4$$.

GP2: first term $$a$$, fifth term $$b$$. Common ratio $$r_2 = (b/a)^{1/4}$$. $$p$$th term = $$a \cdot r_2^{p-1} = a(b/a)^{(p-1)/4}$$.

Setting equal: $$b^5/a^4 = a(b/a)^{(p-1)/4}$$.

$$b^5/a^4 = a \cdot b^{(p-1)/4}/a^{(p-1)/4} = b^{(p-1)/4} \cdot a^{1-(p-1)/4}$$

Comparing powers of $$b$$: $$5 = \frac{p-1}{4} \Rightarrow p - 1 = 20 \Rightarrow p = 21$$.

Verify powers of $$a$$: $$-4 = 1 - \frac{p-1}{4} = 1 - 5 = -4$$ ✓.

The answer is Option (3): $$\boxed{21}$$.