Step-by-Step Solution:

1. Analyze the first observation: A metal rod M is dipped into a colorless concentrated solution of compound N, and the solution turns light blue.

   • The light blue color in aqueous solution is characteristic of the hydrated copper(II) ion, $[Cu(H_2O)_6]^{2+}$. This strongly suggests that the metal rod M is copper (Cu) and it gets oxidized to $Cu^{2+}$ ions.
   • For this reaction to occur, the metal M (Copper) must be more reactive than the metal cation present in compound N. The reaction is a displacement reaction: $M(s) + [cation ext{ from } N]^{n+}(aq) \rightarrow M^{m+}(aq) + [metal ext{ from } N](s)$ Assuming M is Cu, the reaction is: $Cu(s) + [cation ext{ from } N]^{n+}(aq) \rightarrow Cu^{2+}(aq) + [metal ext{ from } N](s)$
   • Let's examine the options based on the reactivity series (or standard electrode potentials). Copper can displace metals that are less reactive than it. The relevant part of the reactivity series is: $... > Pb > H > Cu > Ag > ...$.
     • A: $AgNO_3$: Silver ($Ag$) is less reactive than copper ($Cu$). So, $Cu$ can displace $Ag$ from $AgNO_3$ solution. The reaction is spontaneous. $Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)$
     • B: $Zn(NO_3)_2$: Zinc ($Zn$) is more reactive than $Cu$. No reaction.
     • C: $Al(NO_3)_3$: Aluminum ($Al$) is more reactive than $Cu$. No reaction.
     • D: $Pb(NO_3)_2$: Lead ($Pb$) is more reactive than $Cu$. No reaction.
   • Based on the first observation alone, compound N must be $AgNO_3$.

2. Analyze the second observation: Addition of aqueous $NaCl$ to the blue solution gives a white precipitate O.

   • The blue solution contains $Cu(NO_3)_2$ and possibly unreacted $AgNO_3$ (since the initial solution was concentrated). So, the ions present are $Cu^{2+}$, $Ag^+$, and $NO_3^-$.
   • When $NaCl$ is added, $Na^+$ and $Cl^-$ ions are introduced.
   • The possible precipitates are $AgCl$ and $CuCl_2$. $CuCl_2$ is soluble in water. $AgCl$ is a well-known white precipitate. $Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s) \text{ (white precipitate O)}$
   • This observation is consistent with the solution containing $Ag^+$ ions, which confirms that N is $AgNO_3$.

3. Analyze the third observation: Addition of aqueous $NH_3$ dissolves O and gives an intense blue solution.

   • This observation involves two simultaneous reactions.
   • First, the white precipitate O ($AgCl$) dissolves in aqueous ammonia to form a soluble, colorless complex, diamminesilver(I) chloride. $AgCl(s) + 2NH_3(aq) \rightarrow [Ag(NH_3)_2]^+(aq) + Cl^-(aq)$
   • Second, the light blue $Cu^{2+}$ ions in the solution react with aqueous ammonia to form the tetraamminecopper(II) complex ion, $[Cu(NH_3)_4]^{2+}$, which has a characteristic deep, intense blue color. $Cu^{2+}(aq) + 4NH_3(aq) \rightarrow [Cu(NH_3)_4]^{2+}(aq) \text{ (intense blue solution)}$
   • This final observation perfectly matches the expected reactions for a system containing both $Ag^+$ and $Cu^{2+}$ ions.

Conclusion:

All three observations consistently point to compound N being $AgNO_3$. The metal M is Cu. The precipitate O is $AgCl$. The intense blue color is due to the $[Cu(NH_3)_4]^{2+}$ complex.

Therefore, the correct option is A.