This is from a textbook. The answer says it is about $4.1697 * 10^{-5}$.
This is in the permutations section of the text, so order matters (and from the way it is worded). Given that $P(A)=\frac{n(A)}{n(S)}$, then:
$n(s)=\frac{52!}{(52-7)!}$ --> the number of ways that 7 cards can be given from a deck of 52, where order matters.
n(A)=1 (it seems to me) since, given any set of seven cards, there is only one way that they can be in ascending order.
Thus, by my calculation, $P(A)=1.48*10^{-12}$.
But, I do not get the answer that is in the back of the book. Am I wrong, or is the text?
In a standard deck, the rank of a card is one of $13$ values, and there are four cards of each rank.
In that case, the number of ways to pick $7$ cards of (strictly) increasing rank is $n(A)=\binom{13}{7}4^7.$
This number gives the result you've been given, $0.000041697.$
Even given your reading - essentially that there are $52$ cards labeled from $1$ to $52$ - the answer should be:
$$\dfrac{\binom{52}{7}}{{}_{52}P_{7}}=\frac{1}{7!}$$
There are $n(A)=\binom{52}{7}$ ordered set of seven cards, not $n(A)=1.$