1. Interpret the expression

   The expansion is of $\left(3^{1/2}+5^{1/8}\right)^n.$

   The general term in the binomial expansion is $T_{r+1}=\binom{n}{r}\left(3^{1/2}\right)^{n-r}\left(5^{1/8}\right)^r,$ where $r=0,1,2,\dots,n$.

2. Simplify the powers

   $T_{r+1}=\binom{n}{r}3^{\frac{n-r}{2}}5^{\frac{r}{8}}.$

   For this term to be an integral term, the exponents of both $3$ and $5$ must be integers.

3. Conditions for integrality

   We need: $\frac{n-r}{2}\in \mathbb{Z}$ and $\frac{r}{8}\in \mathbb{Z}.$

   So,

   • $r$ must be divisible by $8$,
   • and $n-r$ must be even.

   Since any multiple of $8$ is even, $r$ is even. Therefore $n-r$ is even iff $n$ is even.

   Hence:

   • if $n$ is odd, there are no integral terms;
   • if $n$ is even, then integral terms occur exactly when $r$ is a multiple of $8$.

4. Count such terms

   For even $n$, valid values of $r$ are $r=0,8,16,\dots,8\left\lfloor \frac{n}{8}\right\rfloor.$

   Therefore, number of integral terms is $\left\lfloor \frac{n}{8}\right\rfloor+1.$

5. Set this equal to 33

   Given exactly $33$ integral terms, $\left\lfloor \frac{n}{8}\right\rfloor+1=33.$

   So, $\left\lfloor \frac{n}{8}\right\rfloor=32.$

   This gives $32\le \frac{n}{8}<33$ $256\le n<264.$

   Also, $n$ must be even.

6. Find the least value of $n$

   The least even integer in $[256,264)$ is $n=256.$

7. Check options

   • A: $264$ gives $\left\lfloor 264/8\right\rfloor+1=34$ terms, not correct.
   • B: $256$ gives $\left\lfloor 256/8\right\rfloor+1=33$ terms, correct.
   • C: $128$ gives $17$ terms, not correct.
   • D: $248$ gives $32$ terms, not correct.

Therefore, the correct option is B.