In the parallelogram ABCD if the internal bisectors of the angle $\angle B$ and $\angle C$ intersect at the point P, then $\angle BPC$ is equal to:

$\frac{\pi}{2}$

Let P.V. of B, A and C be $\vec{0}, \vec{a}$ and $\vec{c}$ respectively.

Now, $\vec{BP}=\lambda_1(\hat{a}+\hat{c})$ and $\vec{CP}=\lambda_2(\hat{a}-\hat{c})$

$\Rightarrow \vec{BP} . \vec{CP}=\lambda_1 \lambda_2(\hat{a}+\hat{c}) .(\hat{a}-\hat{c})=0$

$\Rightarrow \angle BPC=\frac{\pi}{2}$

Hence (4) is correct answer.