*This a revised and expanded version of my post.*

**No**, it is not complete. For simplicity suppose that $X$ is reflexive and separable in which case Dunford and Pettis integrals coincide. Suppose also that $\mu$ is finite and non-atomic. In this case the space of Pettis integrable functions is complete if and only if $X$ is finite-dimensional as proved by Thomas

> G.E.F. Thomas, *Totally Summable Functions with Values in Locally Convex Spaces*,
Measure Theory (Oberwolfach 1975), Lecture Notes in Math. Vol. **541** (1976), 117–131.

This is based on a result o his which asserts that there exist an absolutely sequence summable sequence $(x_n)_{n=1}^\infty$ in $X$ (reflexivity is not needed) and a sequence $(f_n)_{n=1}^\infty$ in the unit ball of $L_1(\mu)$ such that the vector measure

$$\nu(A) = \sum_{n=1}^\infty \int\limits_A f_n(t)\,{\rm dt}\cdot x_n$$

does not have a Pettis intregrable density.

> G.E.F. Thomas, The Lebesgue-Nikodym Theorem for Vector Valued Radon Measures, *Memoirs. of AMS*, **139**, American Mathematical Society, Providence, 1974.

The sequence $(\sum_{k=1}^n f_k\cdot x_k)$ is Cauchy in $\mathcal{P}$ because $(x_n)_{n=1}^\infty$ is absolutely summable, yet it is not convergent as there is no function $F$ such that

$$\lim_{n\to \infty}\int\limits_A \sum_{k=1}^n f_k(t)\cdot x_k\,{\rm d}t\to \int_A F(t)\,{\rm d}t.$$

It seems to me that Pettis integrable functions form a closed subspace of the space of Dunford integrable functions, hence you may extend the above result as in the case where $X$ is infinite-dimensional, you have an incomplete, closed subspace of a normed space, so the space itself cannot be complete.

*Addendum*. Let me point out that if you want to do some functional analysis with the space of Pettis integrable functions, even though incomplete, it is [barrelled][1].


  [1]: http://www.ams.org/journals/proc/1992-114-03/S0002-9939-1992-1107271-2/S0002-9939-1992-1107271-2.pdf