Which must be true?

We can do a little casework to come up with possible digit sets for $(a,b,c,d)$ , from which we can gather more information. Without loss of generality, let's say $a<b<c<d$ . The smallest possible sum $a+b+c$ is $1+2+3=6$ , so the maximum for $d$ is $7$ . Obviously $(1,2,3,7)$ is the only way to have $7$ as the largest element; if $6$ is the largest element, the only possibility is $(1,2,4,6)$ , and if $5$ is the largest element, the only possibility is $(1,3,4,5)$ . If $4$ is the largest element, then $1+2+3+4=10$ is the largest possible sum, so these are the only three possibilities. The elements of $P$ consist of permutations of these three sets of digits. Now let's consider each statement.

1. Each digit in each set is equally likely to be the last digit, so $\frac{1}{4}$ of $(1,2,3,7)$ , $\frac{3}{4}$ of $(1,2,4,6)$ , and $\frac{1}{4}$ of $(1,3,4,5)$ permutations are even. Each set contains $\frac{1}{3}$ of the elements of $P$ , so $\frac{1}{4}\times\frac{1}{3}+\frac{3}{4}\times\frac{1}{3}+\frac{1}{4}\times\frac{1}{3}=\frac{5}{12}$ of elements of $P$ are even. 1 is false.

2. There are $4!=24$ permutations of each set of digits, so $24\times3=72$ total elements total. 2 is true.

3. $3$ is an element of $2$ out of $3$ sets of digits, so the probability that an element of $P$ contains $3$ is $\frac{2}{3}$ . 3 is false.

4. No set contains both $2$ and $5$ . Including both would require a duplication of some digit. 4 is true.

5. The largest element of $P$ is $7321$ and the smallest is $1237$ , so the range is $7321-1237=6084\neq5976$ . 5 is false.

6. All sets contain $1$ as an element. If $1$ were not included, the smallest possible sum would be $2+3+4+5=14>13$ . 6 is true.

So statements 2, 4, and 6 are true and the answer is $\boxed{246}$