It's Simple

$cos\dfrac{\pi}{4} = \sqrt{\dfrac{1}{2}} , ~ \dfrac{\pi}{4} = x$

$\sqrt{\dfrac{1}{2} + \dfrac{1}{2}\sqrt{\dfrac{1}{2}}} = \sqrt{\dfrac{1 + cosx}{2}} = cos\dfrac{x}{2}$

Similarly,

$\sqrt{\dfrac{1}{2} + \dfrac{1}{2}\sqrt{\dfrac{1}{2} + \dfrac{1}{2}\sqrt{\dfrac{1}{2}}}} = cos\dfrac{x}{4}$

thus it becomes

$P(n) = cos\dfrac{x}{2^{n}}cos\dfrac{x}{2^{n-1}}.... cos{x}$

$lim_{n \to \infty}P(x)$

$P(n) = \dfrac{2sin\dfrac{x}{2^{n-1}}cos\dfrac{x}{2^{n-1}}cos\dfrac{x}{2^{n-2}}..... cos\dfrac{x}{2}}{2sin\dfrac{x}{2^{n-1}}}$

$2sinxcosx=sin2x$

thus we get(multiplying and divivding by 2 n times and applying the above identity in each step)

$lim_{n \to \infty}P(n) = lim_{n \to \infty}\dfrac{sin2x}{2^{n}sin\dfrac{x}{2^{n-1}}}$

$lim_{n \to \infty}P(n) = lim_{n \to \infty}\dfrac{sinx}{2^{n}\dfrac{sin\dfrac{x}{2^{n-1}}}{\dfrac{x}{2^{n - 1}}}\times\dfrac{x}{2^{n -1}}}$

$lim_{n \to \infty}P(n) = \dfrac{sin2x}{2x}$

$\dfrac{2sin\dfrac{\pi}{2}}{\pi} = \dfrac{2}{\pi}$