If there are two A’s possible, then the options for person k in include either receiving the first of the two slips or the second. The probability for determining the first of the two is computed in a manner similar to above except that there is one more A and one less other.
\begin{align*}
P(\text{kth as first}) & = P(\text{LL...LA})\\
& = \frac{13 \cdot 12 \cdot ... \cdot (15-k) \cdot 2}{15 \cdot 14 ... \cdot (16-(k+1)) \cdot (16-k)} = \frac{2 \cdot (15-k)}{15 \cdot 14}
\end{align*}
The probability of getting the second A means exactly one of the previous k-1 selections also picked the other A. There are k-1 ways that this could happen. Computing for one of the options and multiplying by k-1 gives
\begin{align*}
P(\text{kth as second}) & = P(\text{LL...LAA})\\
& = (k-1) \cdot \frac{13 \cdot 12 \cdot ... \cdot (15-k) \cdot 2 \cdot 1}{15 \cdot 14 ... \cdot (16-k) \cdot (15-k)} = \frac{2 \cdot (k-1)}{15 \cdot 14}.
\end{align*}
Adding these two together gives
\begin{gather*}
P(\text{getting an A when there are two}) = \frac{2 \cdot (15-k) + 2 \cdot (k-1)}{15 \cdot 14}\\
= \frac{28}{15 \cdot 14} = \frac{2}{15}.
\end{gather*}
For example, if k = 5,
\begin{gather*}
P(\text{5th as first}) = P(\text{LLLLA}) \\
= \frac{13 \cdot 12 \cdot 11 \cdot 10 \cdot 2}{15 \cdot 14 \cdot 13 \cdot 12 \cdot 11} = \frac{20}{15 \cdot 14}
\end{gather*}
\begin{gather*}
P(\text{5th as second}) = P(\text{LL...LAA}) \\
= 4 \cdot \frac{13 \cdot 12 \cdot \cdot 11 \cdot 2 \cdot 1}{15 \cdot 14 \cdot 13 \cdot 12 \cdot 11} = \frac{8}{15 \cdot 14}.
\end{gather*}
Adding these together yields the general result. So, once again, it doesn’t matter which pick you use since the likelihood of getting an A is the same for all positions.