= Solution
For the specified $1/\sqrt{2E_{\mathbf p}}$ mode normalization, the <canonical anticommutation relations> are
$$
\boxed{\{a_{\mathbf p}^{s},a_{\mathbf q}^{r\dagger}\}
=\{b_{\mathbf p}^{s},b_{\mathbf q}^{r\dagger}\}
=(2\pi)^3\delta^{sr}\delta^3(\mathbf p-\mathbf q),}
$$
and every other <anticommutator> of the $a,b$ operators vanishes. For the <Dirac field>, they imply the <Equal-time canonical anticommutator of a Dirac field>
$$
\{\psi_\alpha(t,\mathbf x),\psi_\beta^\dagger(t,\mathbf y)\}
=\delta_{\alpha\beta}\delta^3(\mathbf x-\mathbf y),\qquad
\{\psi_\alpha(t,\mathbf x),\psi_\beta(t,\mathbf y)\}=0.
$$
To check the normalization, the two sectors contribute $u(\mathbf p)u^\dagger(\mathbf p)$ and $v(\mathbf p)v^\dagger(\mathbf p)$ with opposite spatial Fourier signs. Reverse $\mathbf p$ in the second contribution. The completeness relations then give
$$
\sum_s\left[u_s(\mathbf p)u_s^\dagger(\mathbf p)
+v_s(-\mathbf p)v_s^\dagger(-\mathbf p)\right]
=\left[(\not p+m)+(\not{\tilde p}-m)\right]\gamma^0
=2E_{\mathbf p}I,
$$
where $\tilde p=(E_{\mathbf p},-\mathbf p)$. This cancels the mode denominator and leaves the Fourier representation of the delta function. Same-momentum $u^\dagger v$ orthogonality must not be used in place of this reversed-momentum identity.
For arbitrary spacetime points,
$$
\{\psi_\alpha(x),\bar\psi_\beta(y)\}
=(i\not\partial_x+m)_{\alpha\beta}\Delta(x-y),
\qquad
\Delta(z)=\int\frac{d^3p}{(2\pi)^3\,2E_{\mathbf p}}
\left(e^{-ip\cdot z}-e^{ip\cdot z}\right).
$$
For spacelike $z$, choose a Lorentz frame with $z^0=0$; reversing spatial <momentum> shows $\Delta(z)=0$. Its derivatives vanish there as well, so the field <anticommutators> vanish at spacelike separation. This is fermionic <microcausality>. The anticommutation rules make the occupation of each normalized mode zero or one, implementing the <Pauli exclusion principle>, and make the <normal-ordered> free <Hamiltonian> positive for both particle sectors.
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