Upon further reflection I'm no longer convinced that this is an NP hard problem, but this may be because I'm a bit fuzzy on the nuances of the definition of NP hard.

Although the worst case is always proportional to $p!, this problem still isn't as "hard" as a problem where we simply got the number of non-bagels and "yes/no" about whether they were in the right order. If we just got "yes/no" then once we knew all the digits, every additional guess would at most eliminate one of $p! possible permutations. However, in our case we can sometimes know how many are in the right position. Each time we get lucky and make a guess with one additional fermi, we may be able to reduce the number of potential remaining solutions by more than just 1.

For example, suppose the problem selector chooses a solution that will yield $p-$x fermi and $x pico, on the first guess. The number of follow on guesses needed will always be ($p-1) + ($x-1)!. We need ($p-1) to find the identity of out of order elements and at most ($x-1)! guesses to put the elements in the correct position. We can identify the $x out of place digits by taking a known bagel and replacing each digit in turn. Any guess whose pico count drops by 1 will be replacing an out of order digit. After $p-1 such guesses we will know the identity of all out of place values. To put them in the right order, we need to potentially try all possible orderings where none of the $x elements are in their original position. The first of $x elements may be in ($x-1) locations, the second in any of the ($x-2) remaining and so on. Thus we need to consider ($x-1)! arrangements.

However, we only need to make ($x-1)! guesses to put things in the right order if the number of pico is $x for all but the final "guess" (the guess we make when we know the answer but have never actually guessed that particular permutation). If after Q guesses we get $y additional fermi, then instead of guessing each remaining ($x-1)! arrangement, the number of remaining guesses is the minimum of ($x-1)! - Q and ($x-1)+($x-$y-1)!.

However, as I stated earlier, we only have the possibility of making a more informative guess. If "fermis" come late enough in the guessing process we may not be able to use them to reduce the total number of guesses to less than ceiling($v/$p) - 1 + $p!. Some clarification by someone with a firmer understanding of what exactly is meant by NP-hard would be really appreciated so I can use these terms more appropriately.

Best, beth