How should I begin solving this? I know that for months, there are 12, and 3 people from a small group suppose to have birthdays in the same month.
Do I just multiply $12\times 3 = 36$ people?
Or do I use "$\lceil x \rceil$"?
Your question is a straightforward application of the pigeonhole principle. In its simplest form, applied to the context of your question, the pigeonhole principle states that for $m = 12$ months, if there are $n \ge 13$ people in a group, then there is guaranteed to be a month in which at least two people's birthdays occur.
This makes intuitive sense: if we had $n = 12$ people, then each person could be assigned a different birth month. But if we add in one more person, this thirteenth person would necessarily have a birth month in common with one of the other twelve.
The generalization of this idea is given in the Wikipedia article we linked to:
For natural numbers $k$ and $m$, if $n = km + 1$ objects are distributed among $m$ sets, then...at least one of the sets will contain at least $k+1$ objects.
As this applies to your original question, then, we see that $m = 12$ (the number of months), $n$ is the number of people in the group, and $k = 2$ is the maximum number of people allowed to share a birth month before satisfying the criterion of the problem.
Q: How do you avoid having three people with birthdays in the same month while making your group of people as large as possible?
A: By having two in each month. That makes $24$ people. So in which month was the $25$th one born?
A group of twenty-four people can fail to satisfy the request (how?). What if you ask another person to join the group?
12 month * 2 people + 1 maybe.
The Answer is $25$, So the question is assuming that if there were $12$ people in the same room they are all born in separate months, so you would do $2\times12=24$. Now $2$ people are in each month now add one person because whichever month he is born in will allow there to be $3$ in one month.