List- I shows different radioactive decay processes and List- II...

In $\alpha$ decay mass number decreases by $4$ unit and atomic number decreases by $2$ unit.

In ${\beta}^{-}$ decay mass number does not change but atomic number increases by $1$ unit.

In ${\beta}^{+}$ decay mass number does not change but atomic number decreases by $1$ unit.

So for

\left(P\right){}^{238}92U{\to}^{234}91Pa$.</p><p>Number of alpha particle$= \dfrac{238 - 234}{4} = 1$. Due to alpha particle, atomic number should decrease by$2$but it has decreased by$1$only, therefore an additional${\beta}^{-}$decay is required, then net change in atomic number will be$- 2 + 1 = - 1$. Therefore,$P \to 4$.</p><p>\left(Q\right){}^{214}82Pb{\to}^{210}82Pb$.

Number of alpha particle $= \dfrac{214 - 210}{4} = 1$. Due to alpha particle, atomic number should decrease by $2$, but there is no change in atomic number, which means two additional ${\beta}^{-}$ decay is required, then net change in atomic number will be $- 2 + 2 = 0$. Therefore, $Q \to 3$.

\left(R\right){}^{210}81Tl{\to}^{206}82Pb$.</p><p>Number of alpha particle$= \dfrac{210 - 206}{4} = 1$. Due to alpha particle, atomic number should decrease by$2$, but atomic number has increased by$1$, which means three additional${\beta}^{-}$decay is required, then net change in atomic number will be$- 2 + 3 = 1$. Therefore,$R \to 2$.</p><p>\left(S\right){}^{228}91Pa{\to}^{224}88Ra$.

Number of alpha particle $= \dfrac{228 - 224}{4} = 1$. Due to alpha particle, atomic number should decrease by $2$, but atomic number has decreased by $3$, which means one additional ${\beta}^{+}$ is required, then net change in atomic number will be $- 2 - 1 = - 3$. Therefore, $S \to 1$.