Suppose $f$ is continuous and there exist $a>1$ such that $\sup_{x \in \mathbb{R}}|x|^a|f(x)| < \infty$. Suppose there exist $\Omega>0$ such that $\hat{f}=0$ outside $[-\Omega, \Omega]$. Let $0<\lambda \leq \frac{2\pi}{\Omega}.$ Prove that $$\lambda \sum_{n \in \mathbb{Z}}f(n\lambda)=\int_{\mathbb{R}}f(x)dx.$$ Here $\hat{f}$ is the fourier transform of $f$.
I tried to used the poisson summation formula but i'm stuck and can't proceed to the desire conclusion! Any help or hints would be much appreciated!!!
Yes, it's just Poisson summation. In fact it's a particularly simple application: The idea behind Poisson summation is clear, typically the hard part is getting from a formal Fourier series to convergence of the Fourier series. But here it turns out that we get the Fourier series of a trigonometric polynomial, so convergence is no problem.
First, as noted, different authors define the Fourier transform differently, with the $2\pi$s in different places. So when you state this sort of thing you should state explicitly what definition of $\hat f$ you have in mind, even though you're assuming the reader knows all about the Fourier transform; the result could be true with your version but false with the version the reader has in mind.
$\newcommand\ri{\frac1{\sqrt{2\pi}}\int}$ So, for the record, I'm going to be normalizing things as Rudin does, defining $$\hat f(\xi)=\ri f(t)e^{-it\xi}\,d\xi.$$
Now for your problem. Below we'll be dealing with the Fourier series of a $\lambda$-periodic function; this will be less confusing if we start by saying
Replacing $f$ by a suitable dilate, WLOG $$\lambda=2\pi.$$
So now your hypothesis becomes $\Omega\le1$. Which really doesn't seem right; since we're assuming that $\hat f$ vanishes outside $[-\Omega,\Omega]$ surely the condition should be that $\Omega$ is larger than something, not smaller.
So for now forget that inequality; assume $\hat f$ vanishes outside $[-\Omega,\Omega]$ and we'll see what $\Omega$ works. Define $$F(x)=\sum_{n\in\Bbb Z}f(x+2\pi n).$$Then $F$ has period $2\pi$. The bound on $f$ at infinity shows the series converges uniformly on $[-\pi,\pi]$, so $F$ is continuous and the Fourier coefficients of $F$ are related to the Fourier transform of $f$ by $$\hat F(k)=\ri\hat f(k)\quad(k\in\Bbb Z).$$
Assume now that $$\Omega\ge1.$$Then $$\hat F(k)=0,\quad(k\in\Bbb Z, k\ne0),$$so $F$ is actually a constant; since $F$ is continuous we have $$\sum f(2\pi n)=F(0)=\hat F(0)=\ri\hat f(0)=\frac1{2\pi}\int f.$$
Which is what you wanted in a special case, except for the extra $1/2\pi$; as noted above, I suspect this is just due to the Fourier transform being normalized differently.
Note That's a proof if $\lambda=2\pi$ and $\Omega\ge1$. You still have to work out what the constraint on $\Omega$ should be if $\lambda\ne2\pi$. (Apply the above to a suitable dilate of $f$. Or repeat the above, without assuming $\lambda=2\pi$, instead dealing with the Fourier coefficients of $F$ as a $\lambda$-periodic function.)