Solution:
Let us write $P(x)$ as an infinite series:

$$P(x) = 1 + 8x + 4x^2 + 8x^3 + 4x^4 + 8x^5 + 4x^6 + \cdots$$

Notice the coefficients alternate between $8$ and $4$ starting from $8x$.

Let us group the terms:

$P(x) = 1 + (8x + 4x^2) + (8x^3 + 4x^4) + (8x^5 + 4x^6) + \cdots$

Let us factor each group:

$8x + 4x^2 = 4x(2 + x)$

$8x^3 + 4x^4 = 4x^3(2 + x)$

$8x^5 + 4x^6 = 4x^5(2 + x)$

So,

$$P(x) = 1 + 4x(2 + x) + 4x^3(2 + x) + 4x^5(2 + x) + \cdots$$

Now, factor $4(2 + x)$ out of each term except the first:

$$P(x) = 1 + 4(2 + x)[x + x^3 + x^5 + \cdots]$$

The sum inside the brackets is a geometric series with first term $x$ and ratio $x^2$:

$$x + x^3 + x^5 + \cdots = x(1 + x^2 + x^4 + \cdots) = x \left(\frac{1}{1 - x^2}\right)$$

So,

$$P(x) = 1 + 4(2 + x) \cdot \frac{x}{1 - x^2}$$

Now, plug in $x = 1/7$:

First, compute $1 - x^2 = 1 - (1/7)^2 = 1 - 1/49 = 48/49$

$2 + x = 2 + 1/7 = 15/7$

So,

$$P\left(\frac{1}{7}\right) = 1 + 4 \cdot \frac{15}{7} \cdot \frac{1}{7} \cdot \frac{49}{48}$$

Calculate $4 \cdot \frac{15}{7} = \frac{60}{7}$

So,

$$P\left(\frac{1}{7}\right) = 1 + \frac{60}{7} \cdot \frac{1}{7} \cdot \frac{49}{48}$$

$\frac{60}{7} \cdot \frac{1}{7} = \frac{60}{49}$

So,

$$P\left(\frac{1}{7}\right) = 1 + \frac{60}{49} \cdot \frac{49}{48} = 1 + \frac{60}{48} = 1 + \frac{5}{4} = \frac{9}{4}$$

Final Answer:

$$P\left(\frac{1}{7}\right) = \frac{9}{4}$$