Your numerator makes it seem like there is a distinction between the "first" chosen suit and the "second" chosen suit and so on... The denominator there is no way of knowing which was "first" or "second", they are all the same. If you want order to matter then you need order to matter for both numerator and denominator.
Your numerator counted the number of ordered five-card hands where very specifically all suits are present among the first four of the cards. If you wanted order to matter in the denominator like it does for the numerator then use $P(52,5)$ instead (i.e. $52\times 51\times 50\times 49\times 48$)... but then your numerator misses the mark in another way as it neglects to account for scenarios where the repeated suit occurs twice within the first four cards and the fifth card is a new suit such as $A\spadesuit A\heartsuit K\spadesuit A\diamondsuit A\clubsuit$
For a correct approach
Count the number of five card hands. Remove the number of five card hands missing at least a particular suit and which suit that was (among possibly missing more). Add back the number of five card hands missing at least two particular suits, subtracting missing at least three particular suits, etc... You can end at that point since it is impossible to be missing four or more suits at a time here (but more complicated scenarios may have you continuing to alternate between adding and subtracting even further)
$$\dfrac{\binom{52}{5}-\binom{4}{1}\binom{39}{5}+\binom{4}{2}\binom{26}{5}-\binom{4}{3}\binom{13}{5}}{\binom{52}{5}}$$
Choose which suit is repeated. Choose the two cards for that suit simultaneously. For each suit isn't repeated (in alphabetical order) choose which card for that suit is used.
$$\dfrac{\binom{4}{1}\binom{13}{2}13^3}{\binom{52}{5}}$$
If you insist on phrasing this such that order matters... then make sure it matters for numerator and denominator...
Choose the positions of the two cards in the hand for the suit that is repeated. Choose the suit for those cards. Choose the rank of the first, and then the rank of the second. Then for the remaining unused positions, choose a remaining suit and a remaining rank to be used. Take the ratio with respect to the number of ordered hands possible.
$$\dfrac{\binom{5}{2}\cdot 4\cdot 13\cdot 12\cdot 3\cdot 13\cdot 2\cdot 13\cdot 1\cdot 13}{52\cdot 51\cdot 50\cdot 49\cdot 48}$$
The problem can of course be generalized to a larger sized hand of cards. The direct approach quickly devolves into "case hell", too many possible cases to keep track of manually, and so should be avoided. Inclusion-exclusion however will still work nicely.
For a $k$-card hand to have all suits present, this would be $1$ minus at least one suit being missing. Applying inclusion-exclusion over the events of a suit being missing we are left with:
$$1-\left(\dfrac{\binom{4}{1}\binom{39}{k}-\binom{4}{2}\binom{26}{k}+\binom{4}{3}\binom{13}{k}}{\binom{52}{k}}\right)$$
noting that if $k>n$ we treat $\binom{n}{k}$ as zero and that it is impossible to have all four suits simultaneously not appearing (so long as our hand is size at least one)