Since you want ZF+X to prove P and ZF+$\neg$X to prove $\neg$P, it follows that X is equivalent to P. So basically, you are asking for an arithmetic assertion that is independent of ZF, but which satisfies your requirement that it "not follow from the assumption that ZF is sound".

Of course, there are plentiful arithmetic assertions that are independent of ZF, provided that ZF is consistent, such as the Rosser sentence for ZF. But presumably this doesn't satisfy your requirement.

If large cardinals are consistent, then we get more examples. For example, Con(ZF + there is a measurable cardinal) is thought by many set theorists to be true, and in this case it is not provable in ZF, since it implies Con(ZF). But one cannot deduce that it is true just from the assumption that ZF proves only true arithmetic assertions, since if ZF is consistent, then it is consistent with ZF that this assertion is false, yet ZF has an $\omega$-model and hence is arithmetically sound. (This is because the assumption that ZF has an $\omega$-model has weaker consistency strength than a measurable cardinal.)

This reasoning shows that for any consistent theory T whose consistency strength is stronger than an $\omega$-model of ZF (and this includes all of the usual large cardinals), then Con(T) will be an example of a statement independent of ZF yet not deducible merely from the fact that ZF is arithmetically sound.

Since you want ZF+X to prove P and ZF+$\neg$X to prove $\neg$P, it follows that X is equivalent to P. So basically, you are asking for an arithmetic assertion that is independent of ZF, but which satisfies your requirement that it "not follow from the assumption that ZF is sound".

Of course, there are plentiful arithmetic assertions that are independent of ZF, provided that ZF is consistent, such as the Rosser sentence for ZF. But presumably this doesn't satisfy your requirement.

If large cardinals are consistent, then we get more examples. For example, Con(ZF + there is a measurable cardinal) is thought by many set theorists to be true, and in this case it is not provable in ZF, since it implies Con(ZF). But one cannot deduce that it is true just from the assumption that ZF proves only true arithmetic assertions, since if ZF is consistent, then it is consistent with ZF that this assertion is false, yet ZF has an $\omega$-model and hence is arithmetically sound. (This is because the assumption that ZF has an $\omega$-model has weaker consistency strength than a measurable cardinal.)

This reasoning shows that for any consistent theory T whose consistency strength is stronger than an $\omega$-model of ZF (and this includes all of the usual large cardinals), then Con(T) will be an example of a statement independent of ZF yet not deducible merely from the fact that ZF is arithmetically sound.

Update. You have explained that you want to find an arithmetic statement independent of ZF and not provable from the assumption that ZF is sound, where this is formalized in Kelley-Morse set theory.

Indeed, the soundness of ZF is formalizable in KM, since this theory proves the existence of a full satisfaction class for first-order truth. And so in KM we can formulate the assertion $S$ that says "every theorem of ZF is true". Indeed, S is a theorem of KM, which can be seen by induction on the length of proofs, since we have a truth predicate for first-order truth.

So the example sentence $P$ you seek could just be Con(KM). If true, this will be independent of ZF, and you cannot prove it from the assumption that ZF is sound, since then you could prove it in KM, which you cannot by the incompleteness theorem.

(But since you've made the move from ZF to KM, one might think you now want to consider statements P independent of KM, which are not deducible from the "soundness" of KM, and we would be back to the essence of the earlier objections, one level up.)

Let $X$ be the Rosser sentence for ZF, which is independent of ZF if it is consistent, and it is itself an arithmetic assertion.

Since you want ZF+X to prove P and ZF+$\neg$X to prove $\neg$P, it follows that X is equivalent to P. So basically, you are asking for an arithmetic assertion that is independent of ZF, but which satisfies your requirement that it "not follow from the assumption that ZF is sound".

Of course, there are plentiful arithmetic assertions that are independent of ZF, provided that ZF is consistent, such as the Rosser sentence for ZF. But presumably this doesn't satisfy your requirement.

If large cardinals are consistent, then we get more examples. For example, Con(ZF + there is a measurable cardinal) is thought by many set theorists to be true, and in this case it is not provable in ZF, since it implies Con(ZF). But one cannot deduce that it is true just from the assumption that ZF proves only true arithmetic assertions, since if ZF is consistent, then it is consistent with ZF that this assertion is false, yet ZF has an $\omega$-model and hence is arithmetically sound. (This is because the assumption that ZF has an $\omega$-model has weaker consistency strength than a measurable cardinal.)

This reasoning shows that for any consistent theory T whose consistency strength is stronger than an $\omega$-model of ZF (and this includes all of the usual large cardinals), then Con(T) will be an example of a statement independent of ZF yet not deducible merely from the fact that ZF is arithmetically sound.