The operator $a$ is a particle annihilation operator, while $b^{\dagger}$ is an antiparticle creation operator. Acting on the vacuum, $a_{s}(k)|0\rangle=0$, but $b^{\dagger}_{s}(k)|0\rangle\neq0$. In fact, $b^{\dagger}_{s}(k)|0\rangle$ is a one-particle antifermion state (which is not the same as a one-particle fermion state).
The commonality between $a$ and $b^{\dagger}$ is not that they each create a particle. Rather, they each can decrease the fermion number by $1$. (The fermion number is the number of fermions present, minus the number of antifermions—thus zero in the vacuum.) Acting on a one-particle fermion state $a_{s}(k)|k,s\rangle=|0\rangle$, annihilating a fermion with momentum $k$ and spin $s$. The conjugate field $\Psi^{\dagger}$ (or $\bar{\Psi}=\Psi^{\dagger}\gamma_{0}$) involves $a^{\dagger}$, which creates a fermion, and $b$, which annihilates an antifermion. Thus, $\Psi^{\dagger}$ will increase the fermion number by $1$.