Let's add

Relevant wiki: Binomial Coefficient

$\dbinom{2016}{0}+\dbinom{2016}{4}+\dbinom{2016}{8}+\cdots+\dbinom{2016}{2016} \\ = \dfrac{2^{2016}+(1+i)^{2016}+(1-1)^{2016}+(1-i)^{2016}}{4} \\ = \dfrac{2^{2016}+2^{1008}(\cos{504\pi}+i\sin{504\pi})+2^{1008}(\cos{504\pi}-i\sin{504\pi})}{4} \\ = \dfrac{2^{2016}+2^{1009}}{4} = 2^{2014}+2^{1007}. \\ 2^{2014} = (2^{1007})^2 \quad \therefore x = 2^{1007} = a^b \\ \implies a = 2, b = 1007, \quad \therefore a+b = 2+1007 = \boxed{1009}$

Relevant wiki: Binomial Theorem

Similar solution with @Akeel Howell 's

$\begin{aligned} \sum_{k=0}^{2016} \binom {2016}{4n} & = \frac 14 \sum_{k=0}^{2016} \binom {2016}k \left(1^k + i^k + (-1)^k + (-i)^k \right) & \small \color{#3D99F6} \text{Note that }1^k + i^k + (-1)^k + (-i)^k = \begin{cases} 4 \text{ for }k \bmod 4 = 0 \\ 0 \text{ for }k \bmod 4 \ne 0 \end{cases} \\ & = \frac 14 \left((1+1)^{2016} + (1+i)^{2016} + (1-1)^{2016} + (1-i)^{2016} \right) & \small \color{#3D99F6} \text{By binomial theorem} \\ & = \frac 14 \left(2^{2016} + (2i)^{1008} + 0 + (-2i)^{1008} \right) & \small \color{#3D99F6} \text{Note that }(1+i)^2 = 2i \text{ and }(1-i)^2 = -2i \\ & = \frac 14 \left(2^{2016} + 2^{1008}i^{4\times 252} + (-2)^{1008}i^{4\times 252} \right) & \small \color{#3D99F6} \text{Note that }i^4 = 1 \\ & = \frac 14 \left(2^{2016} + 2^{1008} + 2^{1008} \right) \\ & = 2^{-2} \left(2^{2016} + 2^{1009} \right) \\ & = 2^{2014} + 2^{1007} \end{aligned}$

Therefore, $x=2^{1007} \implies a+b = 2+1007 = \boxed{1009}$ .