Consider only one stack. Let $dp[i]$ store a boolean denoting whether the person whose turn it is will win when there are $i$ coins remaining in the stack.

Following are the base cases:

1. $i = 0$ : since the person whose chance it is has no coins to take, he/she loses. Hence, $dp[0] = 0$.
2. $i = 1$ : only one coin, the person whose chance it is can take it and win the game. So, $dp[1] = 1$.
3. $i = k$ : the person whose chance it is can select all the $k$ coins and win since the other cannot take any more. Hence, $dp[k] = 1$.
4. $i = l$ : similarly, $dp[l] = 1$.

Now at each turn a person picks either $1$, $k$, or $l$ coins. To win, one would try to take coins in such a way that with the remaining coins the opponent cannot win. In our case, if either of $dp[i-1]$ , $dp[i-k]$ , $dp[i-l]$ is 0 then $dp[i]$ will be $1$ as we can take that may number of coins such that $dp[rem] = 0$ . But if all of $dp[i-1]$ , $dp[i-k]$ , $dp[i-l]$ are $1$ then in all the three possible options, the the game goes to a state where the opponent wins making $dp[i] = 0$ .
Hence, $dp[i]$ = $!(dp[i-1]\&dp[i-k]\&dp[i-l]).$
Here is my code: LINK