The electric field at point P a distance d from a rod with charges -Q and +Q can be found by integrating the contributions from infinitesimal charge elements on each half of the rod. The contributions from each side are added using the principle of superposition.
To calculate the electric field at point P a distance d from the center of a rod with half charged to -Q and the other half to +Q, one has to consider the contribution of each half separately. For a symmetrically arranged rod, the contributions from opposite charges at the same distance will point in opposite directions because of the difference in sign of the charges. To solve this, you would integrate the contributions to the electric field dE from each infinitesimal piece dx of the rod which contains a charge dq.
Although the problem does not provide the geometry needed to give a numerical answer, the process involves setting up the integral of dE due to a charge element dq = λ dx over the length of each half of the rod, and then using superposition to find the net field by adding the contributions from both sides. The solution will also involve the constant ε0, the vacuum permittivity, as part of the Coulomb's law equation for the electric field.