If $$8 = 3 + \frac{1}{4}(3 + p) + \frac{1}{4^2}(3 + 2p) + \frac{1}{4^3}(3 + 3p) + \ldots \infty$$, then the value of $$p$$ is _______.

$$8 = \sum_{n=0}^{\infty} \frac{3 + np}{4^n} = 3\sum_{n=0}^{\infty}\frac{1}{4^n} + p\sum_{n=0}^{\infty}\frac{n}{4^n}$$

$$\sum_{n=0}^{\infty}\frac{1}{4^n} = \frac{1}{1 - 1/4} = \frac{4}{3}$$

$$\sum_{n=0}^{\infty}\frac{n}{4^n} = \sum_{n=1}^{\infty}\frac{n}{4^n} = \frac{1/4}{(1 - 1/4)^2} = \frac{1/4}{9/16} = \frac{4}{9}$$

$$8 = 3 \times \frac{4}{3} + p \times \frac{4}{9} = 4 + \frac{4p}{9}$$

$$\frac{4p}{9} = 4$$

$$p = 9$$

The answer is $$\boxed{9}$$.