Each component of $\Psi$ satisfies the Klein-Gordon equation, and so we can write (cf. this PSE post)
$$
\Psi_\alpha(x)=\int\frac{d^3p}{\sqrt{(2\pi)^32E_p}}\Big(a_\alpha(p)e^{-ip\cdot x}+b^\dagger_\alpha(p)e^{+ip\cdot x}\Big)
$$
for some operators $a_\alpha,b_\alpha$. If we now require $\Psi$ to satisfy the Dirac equation, we get the algebraic conditions
$$
(\not p+m)a(p)=(\not p-m)b(p)=0
$$
We solve these as follows. Let $u_s(p)\in \mathbb C^4$ with $s=1,2$ be the two linearly independent solutions to $(\not p+m)u(p)=0$, and let $v_s(p)\in \mathbb C^4$ with $s=1,2$ be the two linearly independent solutions to $(\not p-m)v(p)=0$ (there are two and only two solutions because the matrices $\not p\pm m$ have rank 2, as is easily checked). As $u_s,v_s$ are four linearly independent vectors, they are a basis of $\mathbb C^4$, which means we can expand any other vector as linear combinations of them. Thus, we can write
$$
a(p)=\sum_{s=1,2}b_s(p)u_s(p),\qquad b(p)=\sum_{s=1,2} d_s^\dagger v_s(p)
$$
for some scalar operators $b_s,d_s$. Finally, plugging this back into our previous expression, we get
$$
\Psi_\alpha(x)=\int\frac{d^3p}{\sqrt{(2\pi)^32E_p}}\sum_{s=1,2}\Big(b_s(p)u_s(p)e^{-ip\cdot x}+d_s^\dagger v_s(p) e^{+ip\cdot x}\Big)
$$
as required.
For more details see Srednicki ยง37.
