If your question is about complex line bundles up to an algebraic isomorphism, such bundles are classified by [the group of divisors](http://en.wikipedia.org/wiki/Divisor_(algebraic_geometry) for any algebraic variety. There is also only one definition of the group of divisors for smooth varieties, as in your case.
It's not hard to see that group of divisors of $mathbb P^n $ is generated by a hyperplane $H$. The key observation is that a variety defined by a homogenous polynomial of degree $n$ is rationally equivalent to $n$ copies of $H$ (because you can deform the coefficients of polynomial so that it becomes just $x_1^n = 0$).
Therefore the complex line bundles on any $mathbb P^n$ are classified by integer numbers: $mathcal O, mathcal O(1), mathcal O(2)$ etc.
Update: see also the answer by Evgeny who explains the same things as me, but better.
Grassmannians have a cell decomposition into Schubert_cells where each cell is just an affine space $mathbb A^l$. This particularly simple structure means that divisors up to rational equivalence will be the same as algebraic cycles up to rational equivalence and the same as topological $H^2$. Moreover, it will be generated by cells in complex codimension one.
Now it's an exercise to find out from the definition of Schubert cells that there's only 1 cell with this property. To see it, fix a flag $(F_i), F_iin V$. Then the cell is given by the conditions
$$ text{dim}\, Vcap F_{n-k-1+i} = i quad text{for} 0le ile k$$ for subspaces $V$ of dimension $k$.
Therefore the bundles on $Gr(n, k)$ are also numbered by integers.
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