$\lambda(n)$ is the smallest number such that for any k we have $k^{\lambda(n)}\equiv 1 \bmod n$ for any $k$. suppose we have $b\equiv c \bmod \lambda n$. Then $b=c+x\lambda(n)$ for some integer x. Then we have $k^{c}=k^{b+x\lambda(n)}=k^b*(k^{\lambda(n)})^x\equiv k^b*1^x\equiv k^b \bmod n$