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The answer is no. The strength of the theory "ZFC+ there is no projective nonprincipal ultrafilter" is at most that of an inaccessible cardinal, which is strictly weaker than PD.

The reason is that if there is an inaccessible cardinal, then in the Solovay model, the $L(\mathbb{R})$ of the Levy collapse $V[G]$, every set is Lebesgue measurable. It follows that every projective set in $V[G]$ is Lebesgue measurable, and consequently, there can be no nonprincipal projective ultrafilters in $V[G]$, since the existence of a nonprincipal ultrafilter implies the existence of a non-measurable projective set, namely, the ultrafilter itself, as I explain in my answer to a remark of Connes.

Since an inaccessible cardinal has weaker consistency strength than PD, which is equiconsistent with infinitely many Woodin cardinals the assertion that every real is in an inner model with an arbitrarily large finite number of Woodin cardinals (see Andres's comment), we cannot make a model of PD this way (unless our theories are inconsistent).

Perhaps there may be a way to get rid of the inaccessible, by using the property of Baire instead of Lebesgue measurability..measurability. [Update:] And indeed, by a result of Shelah, it is equiconsistent with ZFC that every projective set has the property of Baire, and in that model, there can be no projective nonprincipal ultrafilters, since they would need to be meager and this leads to a contradiction. (See comments by Asaf and Andreas.) So the theory "ZFC + there is no projective nonprincipal ultrafilter" is equiconsistent with ZFC.

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The answer is no. The strength of the theory "ZFC+ there is no projective nonprincipal ultrafilter" is at most that of an inaccessible cardinal, which is strictly weaker than PD.

The reason is that if there is an inaccessible cardinal, then in the Solovay model, the $L(\mathbb{R})$ of the Levy collapse $V[G]$, every set is Lebesgue measurable. It follows that every projective set in $V[G]$ is Lebesgue measurable, and consequently, there can be no nonprincipal projective ultrafilters in $V[G]$, since the existence of a nonprincipal ultrafilter implies the existence of a non-measurable projective set, namely, the ultrafilter itself, as I explain in my answer to a remark of Connes.

Since an inaccessible cardinal has weaker consistency strength than PD, which is equiconsistent with infinitely many Woodin cardinals the assertion that every real is in an inner model with an arbitrarily large finite number of Woodin cardinals (see Andres's comment), we cannot make a model of PD this way (unless our theories are inconsistent).

Perhaps there may be a way to get rid of the inaccessible, by using the property of Baire instead of Lebesgue measurability...

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The answer is no. The strength of the theory "ZFC+ there is no projective nonprincipal ultrafilter" is at most that of an inaccessible cardinal, which is strictly weaker than PD.

The reason we shouldn't expect a converse implication is that if there is merely an inaccessible cardinal, then in the Solovay model, the $L(\mathbb{R})$ of the Levy collapse $V[G]$, every set is Lebesgue measurable. It follows that every projective set in $V[G]$ is Lebesgue measurable, and consequently, there can be no nonprincipal projective ultrafilters in $V[G]$, since the existence of a nonprincipal ultrafilter implies the existence of a non-measurable projective set, namely, the ultrafilter itself, as I explain in my answer to a remark of Connes.

Since an inaccessible cardinal has weaker consistency strength than PD, which is equiconsistent with infinitely many Woodin cardinals, we cannot make a model of PD this way (unless our theories are inconsistent).

Perhaps there may be a way to get rid of the inaccessible, by using the property of Baire instead of Lebesgue measurability...

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