What is the point of path integral for boson and fermion?
The fields that enter into the bosonic path integral are just commuting $c$-number fields - in other words, the field values are just plain old complex numbers, not quantum operators. So you don't need to keep track of the ordering of the fields, which makes them much easier to work with.
Could you please tell me the difference between the two kind of path integrals?
The fields in the bosonic path integral are just plain old complex-number functions on spacetime. The fields in the fermionic path integral are Grassmann-number-valued. When you reverse the order of multiplication of two Grassmann numbers, the formal value of the product changes by a minus sign. So in this case the ordering of the fields does matter, but only up to a sign, so you don't need to muck around with complicated commutators or anticommutators. So while not quite as easy to work with as bosonic fields, they're still a lot easier to work with than quantum-operator-valued fields are.
By the way, you never need to use the path integral - it's mathematically equivalent to the canonically-quantized Hamiltonian formalism with operator-valued fields, which looks a lot more like what you do in nonrelativistic QM. The path integral formalism's just much more mathematically convenient to work with, although admittedly more abstract and less physically intuitive (especially in the fermionic case). (Also, the path integral formalism's manifestly Lorentz invariant, while the Lorentz invariance of the canonical Hamiltonian formalism is much harder to see, because the Hamiltonian transforms under Lorentz transformations as the $T_{00}$ component of the rank-two stress-energy tensor, rather than as a Lorentz scalar like the Lagrangian density.)