Let's denote the left side of inequality that should be proven as $S$. Let $x_{14}x_1$ be the smallest product among all pairwise products of adjacent numbers $x_kx_{k+1}$, $k = 1, 14$ (cyclically adjacent numbers: $x_{15} = x_1$). Then we can bound from above all the summands containing this pair like this:
$$x_{12}x_{13}x_{14}x_1 \le x_{12}x_{13}x_6x_7, \quad x_{13}x_{14}x_1x_2 \le x_{13}x_7x_8x_2, \quad x_{14}x_1x_2x_3 \le x_8x_9x_2x_3,$$
But then
$$\begin{aligned} S &= x_1x_2x_3x_4 + x_2x_3x_4x_5 + \dots + x_{11}x_{12}x_{13}x_{14} + x_{12}x_{13}x_{14}x_1 + x_{13}x_{14}x_1x_2 + x_{14}x_1x_2x_3 \\ &\le x_1x_2x_3x_4 + x_2x_3x_4x_5 + \dots + x_{11}x_{12}x_{13}x_{14} + x_{12}x_{13}x_6x_7 + x_{13}x_7x_8x_2 + x_8x_9x_2x_3 = P. \end{aligned}$$

In every group of $4$ summands from $P$ every multiplier has different remainder modulo $4$, that's why
$$\begin{aligned} & (x_1 + x_5 + x_9 + x_{13})(x_2 + x_6 + x_{10} + x_{14})(x_3 + x_7 + x_{11})(x_4 + x_8 + x_{12}) \le \\ & \le \frac{1}{4^4} \left( \sum_{k=1}^{14} x_k \right)^4 = \frac{1}{4^4}, \end{aligned}$$
what was to be demonstrated.