With the current edit, the game goes like this:
There are $N$ matchsticks on a table, two players (Left and Right) and a pre-ordained number $p$. Play alternates between Left and Right, with Left starting, and on a player's turn that player may remove up to $p$ matchsticks from the table. The person who takes the last matchstick loses.
So you ask how Left can guarantee a win. The answer is that he can't always guarantee a win. For example, suppose $p = 1$ and there are 3 matchsticks. He picks up 1, Right picks up 1, and left picks up the last one, losing.
But Left can always win unless $N = kp + (k + 1)$ for $k \geq 0$. Why is this?
If $N = 1$, Left loses. Ok. If $N = 2$, Left picks one up. If $N = 3$, Left picks 2 up. This works until $N = p + 1$, when Left picks up as many as he can take, $p$. At $p + 2$, no matter how many Left picks up, there are a number of sticks that allow Right to win (Right just has to use the strategy that Left would have used). But at $p + 3$ sticks, all that Left has to do is pick up $1$, landing Right in the losing position $p+2$. This works until $2p+2$, when Left can pick up exactly $p$. At $2p + 3$, Left loses as no matter what move Left does, Right can use the strategy we described above for Left.
For example, suppose $N = 14$ and $p = 5$. Then Left wants to move the game to a 'losing position' $kp + (k + 1)$. Here, that means moving to $2(5) + (2 + 1) = 13$, by picking exactly one up. No matter how many Right picks up, Left's next move will be to $1(5) + (1 + 1) = 7$ (i.e. after Left moves, there will only be $7$ sticks on the table). As this is $6$ away from $13$, Right can't move there. Left's next move will be to exactly $1$ stick on the table.
So $N = kp + (k+1)$ are the 'losing positions' of the game.