What is probability that at least $2$ people have same birthday from group of $N$ people? Question is not that simple.  

There are also leap year included.Leap year will be $366$ days and normal year will be $365$ days.

There is a statement in question that : there are exactly  $\lfloor{\dfrac{n}{4}}\rfloor$ people who born in leap year. 

Now we have to find probability of at least $2$ people having same birthday from group of $N$ people .

Edit : sorry i forgot a word to add!

Any help or hints and calculation and edit in questions are appreciated. Thanks..
Let's rephrase the question. 


  Find the probability that any $2$ people have their birthday in the same day of the year in a group $4n$ people, of which $n$ born were in the leap year and $3n$ people were born in the regular year. 


The complement of the event "any 2 people have their birthday" is "all people have birthday on different days of the year". 

Consider two mutually exclusive possibilites: $\mathcal{A}_1$ - one of these birthdays falls on "Feb. 29", and $\mathcal{A}_2$ -- none of these birthdays fall on the leap day.

The number of configurations of the second kind is
$$
   N\left(\mathcal{A}_2\right) = \prod_{k=1}^{4 n} \left( 365 - (k-1) \right)
$$
and the number of configurations of the first kind is
$$
   N\left(\mathcal{A}_1\right) = n \prod_{k=1}^{4 n-1} \left( 365 - (k-1) \right)
$$
The total configuration size is $N_T=366^n 365^{3n}$, hence the probability of having no birthday duplicates is 

$$
   p_c = \frac{ N\left(\mathcal{A}_1\right) +  N\left(\mathcal{A}_2\right)}{N_T}
$$

Here is a confirmation in Mathematica:

(* computed probability expression *)
p[n_Integer] = (Product[365 - (k - 1), {k, 1, 4 n}] + 
     n Product[365 - (k - 1), {k, 1, 4 n - 1}])/(366^n 365^(3 n));

(* simulation result *) 
In[45]:= With[{n=7,reps=10^8},
   Sum[Boole[DuplicateFreeQ[Join[
       RandomChoice[Range[1,366],n],
       RandomChoice[Range[1,365],3n]]]],{k,1,reps}]/reps]//N

Out[45]= 0.34599694

In[53]:= N[p[7]]

Out[53]= 0.3460041650022472

(* verify that computed value belongs to the sampling 
    confidence interval *)
In[55]:= With[{pSample = Out[45], pExpected = p[7]},
 IntervalMemberQ[
  Interval[pSample + {-1, 1} Sqrt[(pSample (1 - pSample))/10^8]], 
  pExpected]]

Out[55]= True
Suppose we find the negation.
$p'(n)$ - is the probability when there is no two person with equal birthday.
$$p'(n) = (1-\frac{1}{365})(1 - \frac{2}{365})\cdots (1-\frac{n}{365}) = \frac{365!}{365^{n}(365-n)!}$$ (suppose we choose we choose one person from group, then choose another one with different birthday and continue this before n)

So the answer is: $$1-p'(n)$$ and it's ~$\frac{1}{2}$, when $n=23$

UPD : the probability of choosing person with leap year birthday is $s(n) = \frac{\lfloor  {\frac{N}{4}}\rfloor }{N}$ so the total probability will be :
$$(1-s(n))p_{365}(N-\lfloor \frac{N}{4}\rfloor) + s(n)p_{366}(\lfloor \frac{N}{4}\rfloor)$$ where $p_{365}(n)$ is probability for normal year