Chem ionic equation

You're correct in saying that any MHCO₃ is typically solid at room temperature, and at quite high temperatures also - ionic bonds! The only situation it would "split into its ions" is if you dissolved it in water for example.
The question in its entirety is about thermal decomposition - i.e. you can heat a compound and the energy will be used to break covalent bonds. This is exactly what's happening here.
Your solid, MHCO₃, is made from an M⁺ ion and a HCO₃^- ion. Upon heating, it is the covalent bonds in the HCO₃^- ion that break/shift, and a bit of rearrangement of the ionic lattice (i.e. to balance the charges and keep the product compound neutral).
The question is asking for what happens to the HCO₃^- ion only.
This is a thermal decomposition reaction. You're heating the compound to decompose it into different products. The general rule is that HCO₃^- will decompose into CO₃^-, H₂O and CO₂, with some equation balancing (this array of products isn't hard to deduce after staring at the formula for a while, either).
Happy balancing!

You're correct in saying that any MHCO₃ is typically solid at room temperature, and at quite high temperatures also - ionic bonds! The only situation it would "split into its ions" is if you dissolved it in water for example.
The question in its entirety is about thermal decomposition - i.e. you can heat a compound and the energy will be used to break covalent bonds. This is exactly what's happening here.
Your solid, MHCO₃, is made from an M⁺ ion and a HCO₃^- ion. Upon heating, it is the covalent bonds in the HCO₃^- ion that break/shift, and a bit of rearrangement of the ionic lattice (i.e. to balance the charges and keep the product compound neutral).
The question is asking for what happens to the HCO₃^- ion only.
This is a thermal decomposition reaction. You're heating the compound to decompose it into different products. The general rule is that HCO₃^- will decompose into CO₃^-, H₂O and CO₂, with some equation balancing (this array of products isn't hard to deduce after staring at the formula for a while, either).
Happy balancing!

It does tell you in the question what the products should be - there isn’t actually a need to recall the general rule.
I’d personally start the question by setting up an unbalanced equation. You already know the only reactant is HCO3^- (s) and that your products are CO3^2- (s), CO2 (g) and H2O (g).
HCO3^- (s) —> CO3^2- (s) + CO2 (g) + H2O (g)
You can readily deduce that the LHS needs to be doubled up by observing that the RHS has two hydrogens (or that the total charge is -2), but the LHS in the unbalanced equation features only half this charge/number of hydrogens.
2HCO3^- (s) —> CO3^2- (s) + CO2 (g) + H2O (g)
By observation, you can now see that the total numbers of each element on either side are now equal and that the charges balance out, too. That is to say the equation is now balanced.
I would also note that because the question specifies water vapour, it is good practice to include state symbols. Had it just said water and not given any indication of a requirement for state symbols, I wouldn’t bother.