${ }_{88} \mathrm{R}_{\mathrm{a}}^{226}$ is converted into ${ }_{82} \mathrm{P}_{\mathrm{b}}^{206}$ by emission of alpha $(\alpha)$ and beta $(\beta)$ particles. The number of alpha and beta particles emitted are respectively

The nuclear decay from $_{88}\mathrm{Ra}^{226}$ to $_{82}\mathrm{Pb}^{206}$ involves emission of $x$ alpha particles ($_{2}\mathrm{He}^{4}$) and $y$ beta particles ($_{-1}\mathrm{e}^{0}$).

Mass number conservation gives $226 = 206 + 4x$, reducing to $4x = 20$ and $x = 5$.

Atomic number conservation yields $88 = 82 + 2x - y$. Substituting $x = 5$ gives $88 = 82 + 10 - y$, so $y = 92 - 88 = 4$.

The final answer is $\boxed{\text{A}}$.