Why is the force not shown when the body is split into components? In the given figure, for ease of determining the reactions, the body has been split. The component figures shows internal forces but not the load P. How is this possible ?
(Though the figure does not show the components being split at point C, the text mentions so.)

 A: It's just because force $P$ acts at point $C$. It doesn't act on the sections shown in the figure on the right. When drawing free body diagrams, you only draw forces acting on the body. You don't include forces that are relevant to the system that are not acting on the body. In this case you have two free body diagrams shown; one for each section. Force $P$ doesn't act on either section, so it isn't shown.
For another example, let's say I have a block sitting on a table in a rising elevator. Of course there normal force between the table and the elevator floor is responsible for the upward acceleration of the block-table system. But if I were to draw the free body diagram for the block then I wouldn't include that normal force. I would just include the forces acting on the block, which would be its weight and the normal force between the block and the table.
In your case the acceleration of the entire system will determine what the internal forces should be. But just because the internal forces depend on $P$ does not mean it is included in free body diagrams for sections that do not include point $C$.
A: Look at the this figure

So obviously you miss something 
if you calculate the sum of the forces from the left figure you get:
$$R_A-p-R_B=0\tag 1$$
now calculate the sum of the forces from the right figure:
$R_A-p_1=0\quad,p_1-p-p_2=0\quad\,p_2-R_B=0$
thus:
$p_1=R_A$
$p_2=R_B$
thus 
$$p_1-p-p_2=0=R_A-p-R_b=0\tag 2$$
you get the same results , equation (1) is equal equation (2) as it should be
if you do the same calculation from your left and right figures, you don't get
the same results, so it is something wrong. 
