More fun in 2016, Part 2

Let $f(x) = (1+x)^{2016}$ .

Let $\omega = i$ , the primitive fourth root of unity.

Thus it is easy to see that $\begin{aligned} f(1) + f(\omega) + f(\omega^2)+f(\omega^3) & = 4\sum{2016 \choose 4k} + (1 + \omega + \omega^2 + \omega^3)\sum{2016 \choose 4k+1} + (2+2\omega^2)\sum{{2016 \choose 4k+2}}+(1 + \omega + \omega^2 + \omega^3)\sum{2016 \choose 4k+3} \\ & = 4\sum{2016 \choose 4k} = 4P \\ 2^{2016} + (1+i)^{2016} + (1-i)^{2016} & = 4P \\ 2^{2016} + (2i)^{1008} + (-2i)^{1008} & = 4P \\ 2^{2016} + 2^{1009} & = 4P \implies P = 2^{2014} + 2^{1007} \implies 2014 + 1007 = \boxed{3021} \end{aligned}$