Cyclopropane is a highly strained molecule with bond angle of $60^\circ$. The normal tetrahedral bond angle is $\pu{109^\circ{28}'}$, so we'd expect a ring strain of $49.28^\circ$!

For a cyclopropane C atom, the four hybrid orbitals are not equivalent. The two orbitals directed to the outside bonds have more $\ce{s}$- character than a normal $\ce{sp^3}$ orbital, while the involved in ring bonding have less, because the more $\ce{p}$ like they are, the more they resemble the ordinary $\ce{p}$ orbitals whose ordinary bond angle is $90^\circ$. Since the angle strain is a measure of the difference between the preffered angle and the real angle of $60^\circ$, this additional $\ce{p}$ character relieves the strain.  The external bonds have $\sim 33\%$ $\ce{s}$ character, so they are $\sim \ce{sp^2}$ orbitals, while the internal orbitals have $\sim 17\% \ce{~s}$ character so they are $\sim \ce{sp^5}$  orbitals. Thus, each C-C bond is formed by $\ce{sp^5-sp^5}$ verlap. 

The bonds in cyclopropane are called **bent bonds** and they are intermediate in character between $\sigma$ and $\pi$. 

Now, coming to the stability of cyclopropyl methyl carbocation, it is *symmetrically stabilized* by both $\ce{C-C \sigma}$ (2-3 and 2-4) bonds.  You may call it *bent bond resonance*  with the vacant $\ce{p}$ orbital of the carbocation. 

[![enter image description here][1]][1]


It is worth mentioning that cyclopropyl group stabilizes an adjacent postive charge even *better than a phenyl group*. 

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**Source:** **March's Advanced Organic Chemistry**
  [1]: https://i.sstatic.net/eqbIs.png