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No, it is not complete. For simplicity suppose that $X$ is reflexive in which case Dunford and Pettis integrals coincide. Suppose also that $\mu$ is finite.

The (uncompleted) injective tensor product $L_1(\mu)\odot_{\varepsilon} X$ can be isometrically embedded into the space of all Pettis integrable functions $\Omega\to X$ (See Proposition 3.13 in Ryan's Introduction to tensor products of Banach sapces) and all simple functions can be regarded as elements of this tensor product. However if $L_1(\mu)$ is infinite-dimensional, then $L_1(\mu)\odot_{\varepsilon} X$ is complete if and only if $X$ is finite-dimensional, so the space of Pettis integrable functions is complete only in this case.

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.

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.

No, it is not complete. For simplicity suppose that $X$ is reflexive in which case Dunford and Pettis integrals coincide. Suppose also that $\mu$ is finite.

The (uncompleted) injective tensor product $L_1(\mu)\odot_{\varepsilon} X$ can be isometrically embedded into the space of all Pettis integrable functions $\Omega\to X$ (See Proposition 3.13 in Ryan's Introduction to tensor products of Banach sapces) and all simple functions can be regarded as elements of this tensor product. However if $L_1(\mu)$ is infinite-dimensional, then $L_1(\mu)\odot_{\varepsilon} X$ is complete if and only if $X$ is finite-dimensional, so the space of Pettis integrable functions is complete only in this case.

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.

No, it is not complete. For simplicity suppose that $X$ is reflexive in which case Dunford and Pettis integrals coincide. Suppose also that $\mu$ is finite.

The (uncompleted) injective tensor product $L_1(\mu)\odot_{\varepsilon} X$ can be isometrically embedded into the space of all Pettis integrable functions $\Omega\to X$ (See Proposition 3.13 in Ryan's Introduction to tensor products of Banach sapces) and all simple functions can be regarded as elements of this tensor product. However if $L_1(\mu)$ is infinite-dimensional, then $L_1(\mu)\odot_{\varepsilon} X$ is complete if and only if $X$ is finite-dimensional, so the space of Pettis integrable functions is complete only in this case.

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.

No, it is not complete. For simplicity suppose that $X$ is reflexive in which case Dunford and Pettis integrals coincide. Suppose also that $\mu$ is finite.

The (uncompleted) injective tensor product $L_1(\mu)\odot_{\varepsilon} X$ can be isometrically identified with the space of all Pettis integrable functions $\Omega\to X$ (See Proposition 3.13 in Ryan's Introduction to tensor products of Banach sapces). However if $L_1(\mu)$ is infinite-dimensional, then $L_1(\mu)\odot_{\varepsilon} X$ is complete if and only if $X$ is finite-dimensional.

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.

No, it is not complete. For simplicity suppose that $X$ is reflexive in which case Dunford and Pettis integrals coincide. Suppose also that $\mu$ is finite.

The (uncompleted) injective tensor product $L_1(\mu)\odot_{\varepsilon} X$ can be isometrically embedded into the space of all Pettis integrable functions $\Omega\to X$ (See Proposition 3.13 in Ryan's Introduction to tensor products of Banach sapces) and all simple functions can be regarded as elements of this tensor product. However if $L_1(\mu)$ is infinite-dimensional, then $L_1(\mu)\odot_{\varepsilon} X$ is complete if and only if $X$ is finite-dimensional, so the space of Pettis integrable functions is complete only in this case.

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.