The correct order of ionic radii for the ions, P$$^{3-}$$, S$$^{2-}$$, Ca$$^{2+}$$, K$$^+$$, Cl$$^-$$ is:

First, let us note the atomic numbers of the parent atoms: $$\text P\,(Z=15),\; \text S\,(Z=16),\; \text{Cl}\,(Z=17),\; \text K\,(Z=19),\; \text{Ca}\,(Z=20).$$

Now we write the electronic population of each given ion. We subtract or add electrons according to the charge:

$$\begin{aligned} \text P^{3-}: &\; 15+3 = 18 \text{ electrons},\\ \text S^{2-}: &\; 16+2 = 18 \text{ electrons},\\ \text{Cl}^-: &\; 17+1 = 18 \text{ electrons},\\ \text K^+: &\; 19-1 = 18 \text{ electrons},\\ \text{Ca}^{2+}: &\; 20-2 = 18 \text{ electrons}. \end{aligned}$$

So every species possesses exactly $$18$$ electrons. Such a group is called an isoelectronic series.

For an isoelectronic series, we apply the rule: “When the number of electrons is the same, the ionic radius decreases as the nuclear charge $$Z$$ increases.” The reason is that a greater positive charge in the nucleus pulls the same electron cloud more strongly, producing a smaller radius. Symbolically we can write

$$\text{larger } Z \;\Longrightarrow\; \text{larger } Z_{\text{eff}} \;\Longrightarrow\; \text{smaller radius}.$$

Hence we simply arrange the ions in order of increasing atomic number (increasing $$Z$$) to obtain decreasing radius. Starting with the smallest atomic number 15 and moving upward:

$$Z=15 \;(\text P^{3-}) \gt Z=16 \;(\text S^{2-}) \gt Z=17 \;(\text{Cl}^-) \gt Z=19 \;(\text K^+) \gt Z=20 \;(\text{Ca}^{2+}).$$

Thus the descending order of ionic radii is

$$\text P^{3-} \gt \text S^{2-} \gt \text{Cl}^- \gt \text K^+ \gt \text{Ca}^{2+}.$$

When we match this sequence with the listed alternatives, we see that it coincides exactly with Option A.

Hence, the correct answer is Option A.