Assume $n$ keys sorts integers $\epsilon \left \{ 0,1,....(k-1) \right \}$
Then according to counting sort , for $k$ integers can sort $O(n)$ time.
Now, here is the proof by an algorithm:
Say, $L=$ Array of $k$ empty lists for $j$ in $range\left ( n \right )$.
$L\left [ key\left ( A\left [ j \right ] \right ) \right ],append\left ( A\left [ j \right ] \right )$
for $i$ in $range\left [ k \right ]:$
$Output.extend\left ( L\left [ i \right ] \right )$
This algorithm has time complexity $O\left ( n+k \right )$.........$(i)$
Now, using counting sort imagine each integer has base $b.$
Then, $digit=d=\log _{b}k+1$
Now, according to counting sort expression $(i)$ will be $O\left ( n+b \right )$
Then total time$=O\left ( \left ( n+b \right ).d \right )$
$=O\left ( \left ( n+b \right ).\log _{b}k \right )$
$=O\left ( \left ( n \right ).\log _{n}k \right )$ when $b\simeq n$
$=O\left ( 3n \right )=O\left ( n \right )$ [when $k=n^{3}$ a polynomial function]
Source:MIT Lecture