b
Given, \(\overrightarrow{ P }+\overrightarrow{ Q }=\overrightarrow{ P }-\overrightarrow{ Q }\)

Comparing the square of magnitudes, we have

\(P ^2+ Q ^2+2 P Q \cos \theta= P ^2+ Q ^2-2 P Q \cos \theta\)

Thus, either \(\cos \theta=0\) or one of the two vectors is a null vector. If \(P\) is null, then \(\overrightarrow{ P }+\overrightarrow{ Q }=\overrightarrow{ P }-\overrightarrow{ Q }\) gives \(\overrightarrow{ Q }=-\overrightarrow{ Q }\), implying both are null, which is absurd.

Also if they are perpendicular, the direction of sum and difference of vectors will be different, so that case is also ruled out. Hence it can happen only if \(Q\) is null vector.