It turns out yes, this is NP-complete. The proof is a lot more boring than I would have hoped for.

Setup of proof: We show that the problem is in NP and NP-hard.

The problem is clearly in NP. A palindrome can easily be verified in polynomial time by checking the last character against the first, the second to last against the second, and so on. If one does not want to consider spaces of capitalization, spaces can be removed easily, and capital letters can easily be made lowercase.

Now we show that the problem is NP-hard with a reduction from 3-partition.

Suppose that *m* and *n* are integers, and *n* is divisible by 3. Suppose that *l*_{1}, *l*_{2}, *l*_{3}, ... , *l*_{n} are integers where $\frac{m}{4}$ < *l*_{i} < $\frac{m}{2}$ $\forall i$. Then determining if *l*_{1}, *l*_{2}, *l*_{3}, ... , *l*_{n} can be arranged into triplets *l*_{i}, *l*_{j}, *l*_{k} such that *l*_{i} + *l*_{j} + *l*_{k} = *m* is NP-complete. We show that for every instance of this problem, there exists an instance of the previously described palindrome problem such that a "yes" for 3-SAT means a "yes" for the palindrome problem, and a "no" for 3-SAT means a "no" for the palindrome problem.

To make such an instance, we name one string *S* and let it equal "aaa ... aaa,aaa ... aaa, ... ,aaa ... aaa", which is a string that consists of *m* a's, followed by a comma, followed by *m* a's, followed by a comma, and continuing this pattern until it ends with *m* a's. This string contains *n* - 1 commas. In other words, the string is *n* groups of *m*-long strings of a's, separated by commas.

Example: *m* = 6, *n* = 4

*S* = "aaaaaa,aaaaaa,aaaaaa,aaaaaa"

This is 4 groups of 6-long strings of a's, separated by commas.

Then we define strings *s*_{i}, 1 $\leq$ *i* $\leq$ *n*, to contain a number of a's equal to the value of *l*_{i}.

We then define strings *b*_{i} = "," 1 $\leq$ *i* $\leq$ *n* - 1.

So to arrange strings *S*, *s*_{i}, and *b*_{i} ($\forall$*i* on which these strings are defined) into a palindrome, we have to place the long string *S* on one side, and try to form triplets from the strings *s*_{i} so that each triplet has exactly *m* a's in it. Then we just put a comma between each triplet. So this specific type of the palindrome problem is exactly the same as a 3-SAT problem.

An actual proof would need more writing, but (I hope) I've made the main idea clear.