1. Total number of distinct 7-digit numbers

We form a 7-digit number using the multiset of digits: $3,3,4,4,4,5,5$

The total number of distinct arrangements is: $\frac{7!}{2!\,3!\,2!}=\frac{5040}{24}=210$

2. Condition for divisibility by 2

A number is divisible by $2$ if its last digit is even.

Among the given digits, the only even digit is $4$. So the number must end in $4$.

3. Count numbers ending in 4

Fix one $4$ in the last place. The remaining 6 digits are: $3,3,4,4,5,5$

The number of distinct arrangements of these 6 digits is: $\frac{6!}{2!\,2!\,2!}=\frac{720}{8}=90$

4. Required probability

Hence, $P(\text{divisible by }2)=\frac{90}{210}=\frac{3}{7}$

5. Option check

$\frac{3}{7}$ corresponds to Option D.

Therefore, the required probability is: $\boxed{\frac{3}{7}}$