Inspired by Calvin Lin - Why do I love 2015 so much 4

$\begin{aligned} \prod_{n=1}^{1008} \frac{2n}{2n-1} & = \frac{2\times 4 \times 6 \times ... 2016}{1\times 3 \times 5 \times ... 2015} \\ & = \frac{2^{2016}(1008!)^2}{2016!} \quad \quad \quad \quad \quad \quad \color{#3D99F6}{\text{Stirling's approximation } n! \approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n} \\ & \approx \frac{2^{2016}\times 2016 \pi \left(\frac{1008}{e} \right)^{2016}}{\sqrt{4032 \pi} \left(\frac{2016}{e} \right)^{2016}} \\ & = \sqrt{1008 \pi} = 56.273 \\ \\ \Rightarrow 36P & = 36 \left \lfloor \prod_{n=1}^{1008} \frac{2n}{2n-1} \right \rfloor = 36 \left \lfloor 56.273 \right \rfloor = 36 \times 56 = \boxed{2016} \end{aligned}$

We have ,

$=\left \lfloor \displaystyle \prod^{1008}_{n=1} \dfrac{2n} {2n-1} \right \rfloor$

$=\left \lfloor \frac{2}{1}×\frac{4}{3}× \frac{6}{5}×\ldots×\frac{2016}{2015} \right \rfloor$

$=\left \lfloor \frac{(2×4×6×\ldots×2016)^{2}}{2016!} \right \rfloor$

$=\left \lfloor 2^{2016}×\frac{1008!^{2}}{2016!} \right \rfloor$

$=\left \lfloor 2^{2016}×\frac{1}{2016 \choose 1008} \right \rfloor$

Now we will apply the asymptotic properties for Central Binomial Coefficient:

${2n \choose n } \sim \frac {4^n}{\sqrt{\pi n}}$

In this case, $n=1008$

$=\left \lfloor 2^{2016}×\frac{1}{\frac{4^{1008}}{\sqrt{1008\pi}}} \right \rfloor$

$=\left \lfloor 2^{2016}×\frac{\sqrt{1008\pi}}{4^{1008}} \right \rfloor$

$=\left \lfloor \sqrt{1008\pi} \right \rfloor=56=P$

$\Rightarrow 36P=36\cdot 56=\boxed{2016}$