Step 1: Identify Element M.
M does not react with dilute HCl to release hydrogen. This places M below Hydrogen in the electrochemical series. A common example fitting the subsequent reactions (forming a complex with KCN) is Copper (Cu).
Step 2: Reaction with KCN.
Copper(II) sulfate reacts with excess KCN. First, it forms Cu(CN)$_2$, which decomposes to CuCN (Cu is reduced to +1 state) and releases cyanogen gas. Then, CuCN dissolves in excess KCN to form the stable complex $K_3[Cu(CN)_4]$.
\[ \text{Complex P} = [Cu(CN)_4]^{3-} \]
Step 3: Reaction with H$_2$S.
The cyano-complex of Copper(I) is extremely stable. When H$_2$S is passed through it, the concentration of free copper ions is insufficient to exceed the solubility product of copper sulfide. Therefore, no precipitate of MS is formed.
Conclusion: The amount of MS formed is 0.