Let $P_1$ be the probability of the first archer winning. The trick is to notice that if the first two shots both miss, then the situation is the same as it was at the start, so the first player's probability of winning is $P_1$ again. So we have:
$$P_1=p_1+(1-p_1)(1-p_2)P_1$$
You can easily solve this for $P_1$.