1. Write the general term

For $\left(5^{1/2}+7^{1/8}\right)^{1016},$ the general term is $T_{r+1}=\binom{1016}{r}\left(5^{1/2}\right)^{1016-r}\left(7^{1/8}\right)^r,$ where $r=0,1,2,\dots,1016$.

So, $T_{r+1}=\binom{1016}{r}5^{\frac{1016-r}{2}}7^{\frac{r}{8}}.$

2. Condition for an integral term

Since $\binom{1016}{r}$ is always an integer, the term will be an integer only when the powers of both $5$ and $7$ are integers.

So we need: $\frac{1016-r}{2}\in \mathbb{Z}$ and $\frac{r}{8}\in \mathbb{Z}.$

3. Simplify the conditions

• $\dfrac{r}{8}\in\mathbb{Z}$ means $r$ must be divisible by $8$.
• If $r$ is divisible by $8$, then it is certainly even, so $1016-r$ is also even (because $1016$ is even).

Hence the first condition is automatically satisfied whenever $r$ is a multiple of $8$.

Therefore, we only need $r=0,8,16,\dots,1016.$

4. Count such values of $r$

Let $r=8k,$ where $k=0,1,2,\dots,\frac{1016}{8}=127.$

So the number of possible values is $127-0+1=128.$

5. Conclusion

Thus, the number of integral terms is $\boxed{128}.$

So the correct option is B.