Assume in certain year the number of credit obligors of a bank who are at risk of default is Poisson distributed with intensity = 10. Each individual obligor may actually default in that year with probability p = 0.2. Assume mutual independence of the obligors. Let X be the number of obligors that default in that year. Show that X is Poisson distributed and determine its intensity.
I am using the below formula and starting with binomial distribution and when $n->{\infty}$ then the binomial distribution converges to poisson but it is not tying up together with the problem entirely $\mathbb{P}\left[ X=k \right]= \sum_{n=0}^{\infty}\mathbb{P}\left[X=k|N=n\right]\cdot\mathbb{P}\left[N=n\right] $
As i said in the comment, it suffices to sum from $n=k$ to $\infty$. So in our summation we can make the substitution $i=n-k$ and use Taylor series expansion for exponential function.
$\mathbb{P}\left[ X=k \right]= \sum_{n=0}^{\infty}\mathbb{P}\left[X=k|N=n\right]\cdot\mathbb{P}\left[N=n\right]$
$=\sum_{n=k}^{\infty}\mathbb{P}\left[X=k|N=n\right]\cdot\mathbb{P}\left[N=n\right]$
$=\sum_{n=k}^{\infty}{n\choose k}p^k(1-p)^{n-k}\cdot\frac{\lambda^ne^{-\lambda}}{n!}$
$=\sum_{i=0}^{\infty}{k+i\choose k}p^k(1-p)^i\cdot\frac{\lambda^{k+i}e^{-\lambda}}{(k+i)!}$
$=\sum_{i=0}^{\infty}{k+i\choose k}p^k(1-p)^i\frac{\lambda^{k+i}e^{-\lambda}}{(k+i)!}$
$=p^k\lambda^ke^{-\lambda}\sum_{i=0}^{\infty}{k+i\choose k}(1-p)^i\cdot\frac{\lambda^i}{(k+i)!}$
$=p^k\lambda^ke^{-\lambda}\sum_{i=0}^{\infty}\frac{(k+i)!}{k!i!}(1-p)^i\cdot\frac{\lambda^i}{(k+i)!}$
$=\frac{(\lambda p)^k}{k!}e^{-\lambda}\sum_{i=0}^{\infty}\frac{1}{i!}(\lambda(1-p))^i$
$=\frac{(\lambda p)^k}{k!}e^{-\lambda}e^{\lambda(1-p)}$
$=\frac{(\lambda p)^k}{k!}e^{-\lambda p}$
So $X \text{~Poisson}(\lambda p)$ as expected.
In your case $\lambda=10$, $p=0.2$ so the requested intensity is $2$.