1. A reaction is redox if oxidation numbers change in the process.

We check each option.

2. Option A:

$NaCl + KNO_3 \to NaNO_3 + KCl$

This is just exchange of ions.

• $Na$ remains $+1$
• $K$ remains $+1$
• $Cl$ remains $-1$
• In $NO_3^-$, $N$ is $+5$, $O$ is $-2$

No oxidation number changes.

So, A is not a redox reaction.

3. Option B:

$CaC_2O_4 + 2HCl \to CaCl_2 + H_2C_2O_4$

This is an acid-salt reaction.

Check oxidation numbers:

• $Ca$ is $+2$ on both sides.
• In $C_2O_4^{2-}$, let oxidation state of each $C$ be $x$: $2x + 4(-2) = -2$ $2x - 8 = -2$ $2x = 6$ $x = +3$

In $H_2C_2O_4$, carbon is still $+3$.

• $Cl$ remains $-1$
• $H$ remains $+1$

No oxidation number changes.

So, B is not a redox reaction.

4. Option C:

$Mg(OH)_2 + 2NH_4Cl \to MgCl_2 + 2NH_4OH$

This is also a double displacement / acid-base type reaction.

• $Mg$ remains $+2$
• In $NH_4^+$, $N$ is $-3$ and remains $-3$
• $Cl$ remains $-1$
• $O$ remains $-2$, $H$ remains $+1$

No oxidation number changes.

So, C is not a redox reaction.

5. Option D:

The intended reaction is:

$Zn + 2AgCN \to 2Ag + Zn(CN)_2$

(There appears to be a typo in the printed product; silver should be formed as $Ag$, not $Ag^+$, for the reaction to balance as a redox displacement.)

Now check oxidation numbers:

• $Zn$ in elemental form: $0$
• In $Zn(CN)_2$, zinc is $+2$

So zinc is oxidized:

$Zn^0 \to Zn^{2+} + 2e^-$

In $AgCN$, the cyanide ion is $CN^-$, so $Ag$ is $+1$. In product, $Ag$ is elemental silver, so oxidation state is $0$.

Thus silver is reduced:

$2Ag^+ + 2e^- \to 2Ag$

Since oxidation and reduction both occur, D is a redox reaction.

6. Conclusion

The redox reaction is:

$\boxed{D}$