> 56 options right?

Right.

> If I calculate 56P2 I get 3080 which seems way wrong to me.

Yup, why calculate 56P2? 56 is the number you wanted, it accounts for the multiple orderings.

WLOG assume 1 and 3 are our first and last digits chosen, leaving only 0, 2, 4, 5, 6, 7, 8, and 9 to choose the last two from. We have 8 choices of digit, and if repetition were allowed we'd have 8^(2) = 64 possible middle two digits. However we aren't allowing repetition, so of course we instead have 8 choices for the first of the middle two digits, then 7 for the second of the two, making 8 * 7 = 56 possible middle two digits without repetition.

Happy to see you checking this with some common sense upper bounds to notice this cannot be the case.