Step-by-Step Solution:

1. Identify the chemical reaction: When aqueous silver nitrate ($AgNO_3$) is added to an aqueous potassium chloride ($KCl$) solution, a precipitation reaction occurs. Silver chloride ($AgCl$), which is sparingly soluble, precipitates out. The overall reaction is: $KCl(aq) + AgNO_3(aq) \rightarrow AgCl(s) + KNO_3(aq)$ The net ionic equation shows the key change in the solution: $Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)$

2. Analyze the conductance before the equivalence point:

   • Initially, the solution contains $K^+$ and $Cl^-$ ions from the dissociation of $KCl$. These ions are responsible for the initial conductance of the solution.
   • As $AgNO_3$ solution is added, $Ag^+$ ions are introduced. These $Ag^+$ ions react with the $Cl^-$ ions in the solution to form solid $AgCl$. This removes $Cl^-$ ions from the solution.
   • For every $Cl^-$ ion removed, a nitrate ion ($NO_3^-$) from the added $AgNO_3$ enters the solution. The $K^+$ ions are spectator ions, so their amount remains constant (though their concentration decreases due to dilution).
   • Essentially, before the equivalence point, $Cl^-$ ions are being replaced by $NO_3^-$ ions in the solution.
   • The change in conductance depends on the ionic mobilities (or limiting molar ionic conductivities, $\lambda^o$) of the ions being replaced ($Cl^-$) and the ions replacing them ($NO_3^-$).
   • The standard limiting molar ionic conductivities at 298 K are:
     • $\lambda^o_{Cl^-} = 76.3 \, S \, cm^2 \, mol^{-1}$
     • $\lambda^o_{NO_3^-} = 71.4 \, S \, cm^2 \, mol^{-1}$
   • Since the ionic conductivity of $Cl^-$ is slightly greater than that of $NO_3^-$, the replacement of the more mobile $Cl^-$ ions with the less mobile $NO_3^-$ ions causes a slight decrease in the overall conductance of the solution. The dilution effect from adding the titrant also contributes to this decrease.
   • Therefore, the plot of conductance versus the volume of $AgNO_3$ will show a line with a small negative slope before the equivalence point.

3. Analyze the conductance at and after the equivalence point:

   • At the equivalence point, all the $Cl^-$ ions have been precipitated as $AgCl$. The conductance of the solution is at its minimum because the solution primarily contains the salt $KNO_3$ and the concentration of other free ions is minimal.
   • After the equivalence point, further addition of $AgNO_3$ solution introduces excess $Ag^+$ and $NO_3^-$ ions into the solution. These ions do not react and their concentration steadily increases.
   • The addition of these extra ions significantly increases the number of charge carriers in the solution.
   • Therefore, the conductance of the solution increases sharply after the equivalence point. The plot will show a line with a steep positive slope.

4. Conclusion and Plot Identification:

   • Combining the analysis from the steps above, the titration curve should have a 'V' shape.
   • The first part of the curve (before the equivalence point) should show a slight decrease in conductance.
   • The second part of the curve (after the equivalence point) should show a sharp increase in conductance.
   • Looking at the given options:
     • (P) shows a continuous decrease.
     • (Q) shows an initial increase, then a slower increase.
     • (R) shows a continuous increase with a change in slope.
     • (S) shows a slight decrease in conductance followed by a sharp increase. This matches our analysis perfectly.

Therefore, the correct plot is (S).