= Solution
Let $A=\mathbf p\cdot\boldsymbol\sigma$ and $E=\sqrt{\mathbf p^2+m^2}$. The <Pauli multiplication identity> gives $A^2=\mathbf p^2I_2$, and the orthonormal two-component <spinors> obey $\sum_s\xi^s\xi^{s\dagger}=I_2$. Taking the <Dirac adjoint> of the supplied <Dirac spinor> therefore gives, directly,
$$
\begin{aligned}
\sum_su^s(\mathbf p)\bar u^s(\mathbf p)
&=\begin{pmatrix}(E+m)I_2&-A\\ A&-A^2/(E+m)\end{pmatrix}\\
&=\begin{pmatrix}(E+m)I_2&-A\\ A&(m-E)I_2\end{pmatrix}
=\boxed{\not p+m}.
\end{aligned}
$$
For the antiparticle deduction, specify the standard normalization $v^{r\dagger}(-\mathbf p)v^s(-\mathbf p)=2E\delta^{rs}$, just as $u^{r\dagger}u^s=2E\delta^{rs}$. \b[Spanning alone does not specify a normalization:] replacing every $v$ by $2v$ preserves the spanning property but multiplies its proposed <fermion spin sum> by four. The standard normalization is the convention implicit in the <mode expansion of a Dirac field> with the standard <canonical anticommutation relations>.
To use completeness correctly, introduce the <Hermitian> <Dirac Hamiltonian>
$$
h(\mathbf p)=\boldsymbol\alpha\cdot\mathbf p+\beta m,
\qquad\alpha^i=\gamma^0\gamma^i,\quad\beta=\gamma^0.
$$
The <Dirac equation> makes $u^s(\mathbf p)$ and $v^s(-\mathbf p)$ its <eigenvectors> with <eigenvalues> $E$ and $-E$. Their mutual orthogonality and spanning property give complementary <orthogonal projectors>
$$
\frac1{2E}\sum_su^s(\mathbf p)u^{s\dagger}(\mathbf p)=\frac{E+h(\mathbf p)}{2E},\qquad
\frac1{2E}\sum_sv^s(-\mathbf p)v^{s\dagger}(-\mathbf p)=\frac{E-h(\mathbf p)}{2E}.
$$
Multiply the second equality by $2E\gamma^0$ on the right and replace $\mathbf p$ by $-\mathbf p$. For the positive-energy <four-momentum> $p=(E,\mathbf p)$,
$$
[E-h(-\mathbf p)]\gamma^0
=E\gamma^0-\boldsymbol\gamma\cdot\mathbf p-m=\not p-m,
\qquad\boxed{\sum_sv^s(\mathbf p)\bar v^s(\mathbf p)=\not p-m}.
$$
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