Recall that if the sum of each row is some constant number $s$ then $s$ is an eigenvalue.
Here you have it that each row have $n-1$ ones and one $k$ (where the $k$ is on the diagonal) so $n-1+k$ is an eigenvalue.
Regarding $k-1$: simply reduce $(k-1)I$ from the diagonal to find that you get the matrix $J$ (all entries are $1$), so clearly the dimension spanned by the rows is $1$ which gives that the eigenspace of the eigenvalue $k-1$ is of dimension $n-1$.
Note: This gives the geometric multiplicity but since $A^{2}$ is symmetric this is the same as the algebraic multiplicity (because $A^{2}$ is diagonalizable).
ADDED: If $k$ is real (and I believe in your case it is) then note those eigenvalues are different since there is no real solution for the equation $k-1=k^2-k+1$ so we found to eigenvalues and one with geometric multiplicity at least $1$ and the other with geometric multiplicity $n-1$ hence the geometric multiplicity of the first eigenvalue is $1$ and we have it that there are $n$ independent eigenvectors so $A^2$ is diagonalizable wich means that algebraic multiplicity = geometric multiplicity