A block of mass $m$ is on an inclined plane of angle $\theta$. The coefficient of friction between the block and the plane is $\mu$ and $\tan \theta >\mu$. The block is held stationary by applying a force $P$ parallel to the plane. The direction of force pointing up the plane is taken to be positive. As $P$ is varied from $P_1=mg(\sin \theta –\mu \cos \theta )$ to $P_2=mg(\sin \theta +\mu \cos \theta )$, the frictional force $f$ versus $P$ graph will look like

As $\tan \theta >\mu$, the block has a tendency to move down the incline. Therefore a force $P$ is applied upwards along the incline. Here, at equilibrium $P+f=mg \sin \theta$

$\Rightarrow f=mg \sin \theta -P$

Now as $P$ increases, $f$ decreases linearly with respect to $P$.
When $P=mg \sin \theta , f=0$.
When $P$ is increased further, the block has a tendency to move upwards along the incline.
Therefore the frictional force acts downwards along the incline.
Here, at equilibrium $P=f+mg \sin \theta$
$\therefore f=P–mg \sin \theta$
Now as $P$ increases, $f$ increases linearly w.r.t $P$.
This is represented by graph (a)