The problem is telling us that $P(1) = 2\cdot P(2)$, $P(2) = 2 \cdot P(3)$, and $P(3) = 2 \cdot P(4)$. This means $$\begin{aligned}P(1) &= 2^3\cdot P(4) \\ P(2)  &= 2^2 \cdot P(4) \\ P(3)  &= 2 \cdot P(4).\\\end{aligned}$$

In the solution, instead of writing each of these separately, we have $P(k) = 2^{4 - k} \cdot P(4)$ for $k = 1, 2, 3, 4$.