
Explanation
In case 1. You can see it is a 2° carbocation (or secondary carbonation) with angle strain of ~25° on cyclopropane
In case 2. You can see it is 3° carbocation (or tertiary carbocation)
with angle strain of ~10° on cyclopropane
It is obvious that case 2 will be major intermediate yielding $\ce{1-chlorobicyclic [3,2,0] heptane}$ as main product that is C.
Edit1: After reading your question more thoroughly I found that I missed one thing about 'exceptional' stability of cyclopropyl methyl carbocation.
The stability that cyclopropyl provides to carbocation is due to interaction of p-orbitals which is not possible in your case because p-orbitals of carbon having '+' charge becomes out of plane with respect to p-orbitals of cyclopropyl.
Edit2: See image below for reference

Then in case of cyclopropyl methyl carbocation the p-orbitals of both carbon containing '+' charge and the carbon on ring are parallel
In your case the p-orbitals of carbon containing '+' charge is perpendicular to plane of cyclopentyl ring but the p-orbitals of carbon on cyclopropyl ring is parallel to plane cyclopentyl ring, so stability is not provided by cyclopropyl ring.
To sum up you can say that the cyclopentyl ring does not let cyclopropyl ring to provide stability to carbocation which makes ring expansion a preferred option.
Edit3 (Bonus): I asked to my professor about this, he told me that there are two conformational isomers of cyclopropyl methyl carbocation. One is bisected conformation and other is perpendicular conformation. The bisected one allows the carbocation to be exceptionally stable but the perpendicular one isn't that much stable (like in your case) due to less interaction of p-orbital and the partial double bond character cannot be achieved.