[Solution] Number the 4 options for each question as $0,1,2,3$. Each person's answer can be written as a triplet $(i, j, k), i, j, k=0,1,2,3$. Thus, all 64 possible answer sets correspond to the 64 nodes (including vertices) of a $3 \times 3 \times 3$ cube divided into 27 unit cubes. The 4 planes of the cube with nodes (differing by the 3rd component) are drawn separately, and the selected nodes are marked as follows:
It is easy to see that no two black points lie on the same grid line, but for any other node, there is a black point on the same grid line. This indicates that when there are 8 students, and their answers are $(0,0,0),(1,1,0),(1,0,1),(0,1,1),(3,2,2),(2,3,2),(2,2,3),(3,3,3)$, all the requirements of the problem are satisfied.

On the other hand, for 7 students, their answer arrays correspond to 7 nodes of the cube, referred to as black points. By the pigeonhole principle, among the 4 planes shown, there is always one plane with at most 1 black point. On this plane, excluding the two grid lines where the black point lies (if any), there are at least 9 nodes. On the other planes, there are at most 7 black points, and each black point's grid line can contain at most 1 of the 9 points. Therefore, it is still possible to select 1 point from the 9 points, such that when the 8th student, represented by this point, joins the first 7, property $P$ still holds. This means that no group of 7 students can meet all the requirements of the problem.
In summary, the minimum number of students required to meet all the requirements of the problem is 8.
6.52 In an exam of a class, there are 3 multiple-choice questions, each with 4 options. It is known that a group of students' answers have property $P$: any two students' answers have at most 1 question with the same answer. Moreover, if this group of students is increased by 1 person, regardless of this person's answers, property $P$ no longer holds. How many students are there in this group at minimum?