Yes, your answer is correct for the question of distributing the 52 cards to thirteen players who each receive four cards apiece such that each receives exactly one card of each suit. (Note that the related question of distributing thirteen cards to four players such that each receives at least one of each suit is far far more complicated)
The mentioned approaches of "For each suit, iterate through the cards and assign which player receives the card. Repeat for each suit" giving $(13\times 12\times 11\times \cdots \times 2\times 1) \times (13\times 12\times \cdots \times 2\times 1)\times (13\times \cdots \times 1)\times (13\times \cdots \times 1) = (13!)^4$ and the approach of "For each player, iterate through the suits and assign a remaining card of that suit to the player. Repeat for each player" giving $(13\times 13\times 13\times 13)\times (12\times 12\times 12\times 12)\times \cdots \times (1\times 1\times 1\times 1) = (13^4)\times (12^4)\times \cdots \times (1^4)$ are both correct and both give the same final value.
These can be seen to be equal by considering that if these were expanded we would have a total of $52$ terms in the overall products of each, and in each we have exactly four $13$'s... exactly four $12$'s... exactly four $11$'s and so on albeit in a different order. But, thanks to the commutativity and associativity properties of multiplication this does not change the final resulting value. Order within a product does not matter.
The concern about player order is unfounded. When talking about "people" in a combinatorics or probability setting, it is generally assumed that each person is distinguishable. With distinguishable objects, we may always assume that there exists some "canonical order" for them. In the case of people, for example, we may assume that they all have different names and we can refer to them in alphabetical order... or for machines we may assume they all have different serial numbers and we can refer to them in order of their serial numbers etc... When distributing the cards to the people in terms of "the first player" vs "the second player" this could have been done in the canonical ordering of the players by name or it could have been done in the canonical ordering of the players by turn order... but the concept of turn order is unnecessary to exist here for the problem to make sense.
If desired, you may include in your problem the concept of turn order and if you say that the turn order of the players has not yet been defined and you wish to consider two ways to deal the cards to be distinct if the turn order was different, then this introduces an additional factor of $13!$, and this is true regardless which approach you had taken to see the original answer. Traditionally however, in problems like this it is assumed that the turn order is already predescribed and the players are already sitting in their respective seats.