How can a left-handed fermion field create a right-handed antifermion?

My question - which is likely stupid or appears due to some confusion - stems from the following considerations: when quantizing canonically we are told (see any book on QFT) that a Dirac fermion field can be expressed as

$$\psi(x) = \int d\tilde{p} (u \cdot\mathbf{a}\,e^{-ipx} + v\cdot\mathbf{b}^\dagger\, e^{ipx})$$

where $\mathbf{a}$ destroys fermions and $\mathbf{b}$ creates antifermions and $u, v$ are Dirac spinors. A left-handed chirality projector $P_L = \frac{1-\gamma_5}{2}$ selects the part of this field $\psi_L$ that transforms according to the left-handed irreducible Lorentz representation.

So, how can the state created by $P_L (v\cdot \mathbf{b}^\dagger)$ be a right-handed antifermion? (as seen for instance from the fact that it interacts weakly)

If I understand correctly, you ask how a so-called "right-handed antiparticle", for example, $\nu^c_R$, can interact via the left-chiral weak interaction.
Confusion originates from whether anti-particle is taken to mean charge-conjugation and parity inversion, $\mathcal{CP}$ or only charge-conjugation, $\mathcal{C}$. Because fermions transform non-trivially under $\mathcal{P}$, care must be taken.
For example, for the left-handed chiral massless neutrino, only the $s=2$ Dirac spinor survives in the field $$\psi_L = \int d \tilde p \left(a_{p,s=2}u^{s=2}_Le^{-ipx}+b_{p,s=2}^\dagger u^{s=2}_Le^{+ipx}\right).$$
The first term destroys a particle with left-handed helicity, $h=-1/2$ and the second term creates an anti-particle with right-handed helicity, $h=+1/2$. We see that the anti-particle in the left-handed chiral neutrino field has right-handed helicity.
This is actually to be expected! The field destroys a particle and creates the $\mathcal{CP}$ anti-particle. The parity, $\mathcal{P}$, operation reverses the helicity (and the chirality). The charge conjugation, $\mathcal{C}$, operation reverses the chirality! Thus the combined $\mathcal{CP}$ operation does not change the chirality.
For the anti-neutrino, we can write, $$(\nu_L)^{CP} = (\nu_R)^{C} = (\nu^{C})_L.$$ The anti-neutrino has right-handed helicity but left-handed chirality and thus feels the weak interaction. To me "right-handed antiparticle" is misleading.