1. Understand the domain and codomain

The set $\{1,3,5,\dots,99\}$ has all odd numbers from $1$ to $99$, so it contains $50$ elements.

Similarly, the codomain $\{2,4,6,\dots,100\}$ also has $50$ elements.

Since $f$ is bijective, it is simply a permutation of the $50$ even numbers.

2. Identify the constrained inputs

We are given $f(3) \ge f(9) \ge f(15) \ge f(21) \ge \cdots \ge f(99).$

The inputs involved are: $3,9,15,21,\dots,99.$

This is an arithmetic progression with first term $3$, common difference $6$, and last term $99$.

Number of terms: $n=\frac{99-3}{6}+1=\frac{96}{6}+1=16+1=17.$

So there are 17 special domain elements whose images must appear in non-increasing order.

3. Count bijections satisfying the order restriction

We must assign distinct even numbers to these 17 inputs.

• First, choose which 17 values from the 50 even numbers will be used for $f(3),f(9),\dots,f(99).$ This can be done in $\binom{50}{17}$ ways.

• Once these 17 values are chosen, because of the condition $f(3) \ge f(9) \ge \cdots \ge f(99),$ and all values are distinct (since $f$ is bijective), their arrangement is forced: they must be placed in strictly decreasing order.

So for the constrained 17 positions, there is exactly $1$ valid arrangement after choosing the values.

• The remaining $50-17=33$ domain elements can be assigned the remaining 33 even numbers in any order: $33!$ ways.

Hence total number of bijections is $\binom{50}{17}\cdot 33!$

Now simplify: $\binom{50}{17}\cdot 33!=\frac{50!}{17!33!}\cdot 33!=\frac{50!}{17!}={}^{50}P_{33}.$

Since ${}^{50}P_{33}=\frac{50!}{(50-33)!}=\frac{50!}{17!},$ this matches the above count.

4. Check options

• A: ${}^{50}P_{17}=\frac{50!}{33!}$ — not correct.
• B: ${}^{50}P_{33}=\frac{50!}{17!}$ — correct.
• C: $33!\times 17!$ — not correct.
• D: $\frac{50!}{2}$ — not correct.

5. Final answer

$\boxed{{}^{50}P_{33}}$

So the correct option is B.