The ionic radii of K$$^+$$, Na$$^+$$, Al$$^{3+}$$ and Mg$$^{2+}$$ are in the order:

We have to arrange the ions $$\text{K}^+ ,\; \text{Na}^+ ,\; \text{Al}^{3+} ,\; \text{Mg}^{2+}$$ in increasing order of their ionic radii. The key idea from chemical periodicity is: for species having the same number of electrons (they are called isoelectronic), a larger nuclear charge ($$Z$$, the atomic number) pulls the electron cloud more strongly toward the nucleus, so the ionic radius becomes smaller. Mathematically one writes

$$Z_{\text{effective}} \propto Z$$

and

$$\text{Greater } Z_{\text{effective}} \; \Longrightarrow \; \text{Smaller radius}.$$

Let us first note the atomic numbers and the total electrons after ionisation:

$$ \begin{aligned} \text{Na}^+ &: Z = 11,\; \text{electrons} = 11 - 1 = 10, \\ \text{Mg}^{2+} &: Z = 12,\; \text{electrons} = 12 - 2 = 10, \\ \text{Al}^{3+} &: Z = 13,\; \text{electrons} = 13 - 3 = 10, \\ \text{K}^+ &: Z = 19,\; \text{electrons} = 19 - 1 = 18. \\ \end{aligned} $$

We see that $$\text{Na}^+ ,\; \text{Mg}^{2+},\; \text{Al}^{3+}$$ all possess $$10$$ electrons, hence they form an isoelectronic series. For this trio the order of $$Z$$ is

$$Z(\text{Al}^{3+}) = 13 \; > \; Z(\text{Mg}^{2+}) = 12 \; > \; Z(\text{Na}^+) = 11.$$

Therefore, applying the earlier rule, their radii follow the reverse order:

$$r(\text{Al}^{3+}) \; < \; r(\text{Mg}^{2+}) \; < \; r(\text{Na}^+).$$

Now let us compare $$\text{Na}^+$$ with $$\text{K}^+$$. These two ions are not isoelectronic: $$\text{Na}^+$$ has $$10$$ electrons (configuration $$1s^2 2s^2 2p^6$$), while $$\text{K}^+$$ has $$18$$ electrons (configuration $$1s^2 2s^2 2p^6 3s^2 3p^6$$). Because $$\text{K}^+$$ possesses electrons in the third shell ($$n = 3$$), its electron cloud is farther from the nucleus than that of $$\text{Na}^+$$ whose electrons reside only up to the second shell ($$n = 2$$). Hence

$$r(\text{K}^+) \; > \; r(\text{Na}^+).$$

Combining both deductions we obtain the full ascending sequence of radii:

$$r(\text{Al}^{3+}) \; < \; r(\text{Mg}^{2+}) \; < \; r(\text{Na}^+) \; < \; r(\text{K}^+).$$

Expressed simply, the order is

$$\text{Al}^{3+} \; < \; \text{Mg}^{2+} \; < \; \text{Na}^+ \; < \; \text{K}^+.$$

Comparing this with the given options, we find that Option C exactly matches this order.

Hence, the correct answer is Option C.