The correct order of ionic radii for the ions P3-, S2-, Ca2+, K+, Cl- is:

🧠 Five ions, all carrying 18 electrons — Z decides the ranking $\ce{P^{3-}}$ ($Z=15$), $\ce{S^{2-}}$ ($Z=16$), $\ce{Cl^-}$ ($Z=17$), $\ce{K^+}$ ($Z=19$), $\ce{Ca^{2+}}$ ($Z=20$). Same 18-electron cloud throughout. The only thing that changes is how many protons squeeze that cloud. Fewer protons → looser cloud → bigger ion.

🗺️ Rank from largest to smallest by increasing Z Start with the lowest Z: $\ce{P^{3-}}$ has only 15 protons holding 18 electrons → biggest. Add one proton at a time: $\ce{S^{2-}}$ (16), $\ce{Cl^-}$ (17), $\ce{K^+}$ (19), $\ce{Ca^{2+}}$ (20) → smallest.

Decreasing size: $\ce{P^{3-}} > \ce{S^{2-}} > \ce{Cl^-} > \ce{K^+} > \ce{Ca^{2+}}$.

⚠️ The trap Option (b) swaps the last two: $\ce{Ca^{2+}} > \ce{K^+}$. Many students think "Ca is below K in size in the periodic table" — that is true for the neutral atoms. But for the ions, $\ce{Ca^{2+}}$ has lost more electrons AND has more protons. So $\ce{Ca^{2+}}$ is the smallest, not $\ce{K^+}$.

$\boxed{\text{Answer: (a)}}$