Products are

The correct option is the one showing cyclohexanol and tert-butyl iodide (represented as a cyclohexane ring with a hydroxyl group, and ${\left(C{H}_3\right)}_{3}C-I$).

Step-by-step Explanation:

1. Understanding the Reactants:
The starting material shown in the question is cyclohexyl tert-butyl ether, which is an unsymmetrical ether. In this molecule, the oxygen atom is connected to:
- A secondary alkyl group (cyclohexyl ring)
- A tertiary alkyl group (tert-butyl group, $-C{\left(C{H}_3\right)}_3$)

2. Protonation of the Ether:
When treated with a strong acid like cold hydroiodic acid ($HI$), the lone pairs on the ether oxygen atom abstract a proton ($H^{+}$) from $HI$ to form a protonated ether (oxonium ion intermediate):

$\text{Ether}+H^{+}\to \text{Protonated Ether (Oxonium ion)}$

3. Determining the Mechanism (S_N1 vs S_N2):
- Generally, the reaction of unsymmetrical ethers with cold $HI$ occurs via an S_N2 pathway, where the iodide nucleophile ($I^{-}$) attacks the less sterically hindered carbon atom.
- However, if one of the alkyl groups attached to the ether oxygen is a tertiary alkyl group (such as the tert-butyl group), the reaction mechanism shifts to an S_N1 pathway.
- This shift occurs because the tertiary C-O bond can easily cleave to generate an exceptionally stable tert-butyl carbocation (${\left(C{H}_3\right)}_3{C}^{+}$).

4. Cleavage and Carbocation Formation:
The protonated ether undergoes heterolytic cleavage at the C-O bond connecting the oxygen and the tert-butyl group:

$\text{Protonated Ether}\to \text{Cyclohexanol}+{\left(C{H}_3\right)}_3{C}^{+}$

This leaves behind cyclohexanol as the alcohol product.

5. Nucleophilic Attack:
The remaining iodide ion ($I^{-}$) rapidly attacks the highly reactive tert-butyl carbocation to form the halide product, tert-butyl iodide:

${\left(C{H}_3\right)}_3{C}^{+}+I^{-}\to {\left(C{H}_3\right)}_{3}C-I$

Therefore, the final products of this reaction are cyclohexanol and tert-butyl iodide.