To have the product divisible by as great power of 2 as possible, each of the factors has to be divisible by the greatest power of 2 possible. By adding an even number to a prime number, the sum is divisible by 2 only if the original prime is even. As the only even prime number is 2, it has to be chosen to all even positions. Their respective sums are 4, 6, 8, and 10, and the respective maximal powers of 2 which they are divisible with are $2^2$, $2^1$, $2^3$, and $2^1$. Let's investigate the odd positions one by one.
As 127 is a prime and $127+1$ is the maximal power of 2 less than $200+1$, the first position is 127. As $125 = 5^3$, the maximal power of 2 dividing the third sum cannot be $2^7$. However, 61 is a prime and $61 + 3 = 2^6$, hence 61 is suitable for the third position. As $123 = 3 \cdot 41$, the maximal power of 2 dividing the fifth sum cannot be $2^7$. However, $59$ is a prime and $59 + 5 = 2^6$, hence $59$ is suitable for the fifth position. As $121 = 11^2$, the maximal power of 2 dividing the seventh sum cannot be $2^7$. The natural numbers smaller than 200 which are divisible by $2^6$ are 64 and 192. The respective numbers which would result in these numbers are $57 = 3 \cdot 19$ and $185 = 5 \cdot 37$, neither of which is a prime. However, 89 is a prime and $89 + 7 = 3 \cdot 2^5$, hence 89 is suitable for the seventh position. In summary, the maximal power of 2 which divides the product of the eight numbers is $2^{7+2+6+1+6+3+5+1} = 2^{31}$.