Peak at m/z = 39 in mass spectrum of 3-chloropropene

I've noticed that in the mass spectrum of 3-chloropropene, there is a peak at m/z = 39, which to me, indicates the formation of the C3H3 ion.

However, given that 3 bonds would have to be broken to form this ion, how can the frequency of this cleavage be explained?

It would indeed be $$\ce{C3H3^+}$$ assuming predominant isotopic species. Although there can be other isomers, the cyclic isomer, the cyclopropenyl cation is strongly stabilized by aromaticity.
As PLD comments, there are really only two steps, not three, to forming this cation. The allyl cation ($$\ce{C3H5^+},m/z=41)$$ is first obtained by loss of the chlorine atom plus an electron. The aromatic cyclopropenyl cation noted above is then obtained by loss of $$\ce{H2}$$.