A beautiful series!

Consider the double-angle cosine identity below:

$\begin{aligned} \cos \theta & = 2\cos^2 \frac \theta 2 - 1 \\ \implies \cos \frac \theta 2 & = \sqrt{\frac {1+\cos \theta}2} & \small \color{#3D99F6} \text{For }\theta = \frac \pi 4 \implies \cos \frac \pi 4 = \frac {\sqrt 2}2 \\ \cos \frac \pi 8 & = \frac {\sqrt{2+\sqrt 2}}2 & \small \color{#3D99F6} \text{Similarly } \\ \cos \frac \pi{16} & = \frac {\sqrt{2+\sqrt{2+\sqrt 2}}}2 & \small \color{#3D99F6} \text{and so on...} \end{aligned}$

This means that the product given is as follows:

$\begin{aligned} P & = \cos \frac \pi 4 \times \cos \frac \pi 8 \times \cos \frac \pi {16} \times \cdots \\ & = \lim_{n \to \infty} \prod_{k=2}^{\color{#3D99F6}n} \cos \frac \pi{2^k} \\ & = \lim_{n \to \infty} \frac {\sin \frac \pi{2^n}\cos \frac \pi{2^n}}{\sin \frac \pi{2^n}} \prod_{k=2}^{\color{#D61F06}n-1} \cos \frac \pi{2^k} \\ & = \lim_{n \to \infty} \frac {\sin \frac \pi{2^{n-1}}}{2 \sin \frac \pi{2^n}} \prod_{k=2}^{n-1} \cos \frac \pi{2^k} & \small \color{#3D99F6} \text{Similarly } \\ & = \lim_{n \to \infty} \frac {\sin \frac \pi{2^{n-2}}}{2^2 \sin \frac \pi{2^n}} \prod_{k=2}^{n-2} \cos \frac \pi{2^k} & \small \color{#3D99F6} \text{and so on...} \\ & = \lim_{n \to \infty} \frac {\sin \frac \pi 2}{2^{n-1} \sin \frac \pi{2^n}} \\ & = \lim_{n \to \infty} \frac 1{2^{n-1} \sin \frac \pi{2^n}} \\ & = \lim_{n \to \infty} \left(\frac \pi 2 \times \frac {\sin \frac \pi{2^n}}{\frac \pi{2^n}} \right)^{-1} & \small \color{#3D99F6} \text{Note that }\lim_{x \to 0} \frac {\sin x}x = 1 \\ & = \frac 2\pi \end{aligned}$

Therefore, the area of the circle with radius $P$ is $\pi \left(\dfrac 2\pi\right)^2 = \dfrac 4\pi \approx \boxed{1.273}$ .