Interesting ratio

$\begin{aligned} A & = \prod_{n=0}^{1000} \exp \dbinom {1000}n = \exp \sum_{n=0}^{1000} \binom{1000}n = \exp \left(2^{1000}\right) \\ B & = \exp \left( 3 \sum_{n=0}^{333} \binom {999}{3n} \right) = \exp \left(2^{999}-2 \right) & \small \color{#3D99F6} \text{See note.} \\ \frac AB & = \exp \left(2^{1000} - 2^{999} + 2 \right) = \exp \left(2^{999}+2\right) \end{aligned}$

$\implies c+d+b = 2+999+2 = \boxed{1003}$

$$

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Note:

$\small \begin{aligned} (1+1)^{3n} & = 1 + \binom {3n}1 + \binom {3n}2 + \binom {3n}3 + \cdots + \binom {3n}{3n} \\ (1+\omega)^{3n} & = 1 + \binom {3n}1 \omega + \binom {3n}2 \omega^2 + \binom {3n}3 + \cdots + \binom {3n}{3n} & \color{#3D99F6} \omega \text{ is the third root of unity.} \\ (1+\omega^2)^{3n} & = 1 + \binom {3n}1 \omega^2 + \binom {3n}2 \omega + \binom {3n}3 + \cdots + \binom {3n}{3n} & \color{#3D99F6} \text{Note that }1+\omega + \omega^2 = 0 \\ 2^{3n} + (1+\omega)^{3n} + (1+\omega^2)^{3n} & = 3 + 3 \binom {3n}3 + 3 \binom {3n}6 + 3\binom {3n}9 + \cdots + 3\binom {3n}{3n} \\ 2^{3n} + (-\omega^2)^{3n} + (-\omega)^{3n} & = 3 \sum_{k=0}^{3n} \binom {3n}{3k} \\ \implies 3 \sum_{k=0}^{3n} \binom {3n}{3k} & = 2^{3n} - 2 \end{aligned}$