If there are random variables $X$ and $N$ which satisfies $$N\sim poisson(\lambda) \\ X\mid N\sim B(N, p)$$ $N$ is a poisson distribution with expectation $\lambda$ and $X$ given $N$ is a binomial distribution with N trials and p success probability.
I already get if $X, Y$ are poisson distribution with expectations $\lambda p$ and $\lambda(1- p), N= X+ Y$ then it satisfies above conditions.
But, I cannot explain to myself why it has to be like that.
Could I find out distribution of $X$ and reasonable explanation for why is it ?
The probability mass function of $X$ can be written as $$ P(X=k) = \sum_{i=0}^{\infty}P(X=k|N=i)P(N=i) $$ You said, $X|N$ is a binomial with parameter $N$ and $p$, therefore for any non-negative integer $k$
$$ P(X=k) = \sum_{i=k}^{\infty} {i\choose k} p^{k}(1-p)^{i-k} \exp(-\lambda)\frac{\lambda^i}{i!} $$ $$= \exp(-\lambda p)\frac{(\lambda p)^k}{k!}\sum_{i=k}^{\infty}\exp(-\lambda(1-p))\frac{(\lambda(1-p))^{i-k}}{(i-k)!}. $$ Now write $j=i-k$ and you will see that the last sum is $1$ and thus
$$ P(X=k) =\exp(-\lambda p)\frac{(\lambda p)^k}{k!} $$ for $k=0,1,\ldots$.