Step 1: Note what a point moment does to a BM diagram.
The shear force diagram changes only at a point load, while the bending moment diagram jumps at a point where an external couple acts. That jump equals the couple itself, $M$.
Since beam PR carries no other load, the shear force stays constant along the whole span.
Step 2: Use symmetry and the end conditions.
Both P and R are simple supports, and a simple support carries zero moment. So the bending moment must be zero right at P and right at R.
Because Q sits exactly at the centre, the beam looks the same from either end. So the moment just left of Q and just right of Q must have equal size, splitting the jump $M$ into two equal halves of $M/2$.
Step 3: Build the diagram from these facts.
Starting from zero at P, the moment must vary linearly up to magnitude $M/2$ just before Q. It then jumps across to the opposite sign, then varies linearly again back to zero at R.
Checking each choice: B and D show no discontinuity at Q at all, just one smooth value, which contradicts a concentrated moment's jump. C shows a moment of $M/2$ surviving at the supports, which a pin or roller cannot sustain.
Final Answer:
Only the diagram with zero end values and a symmetric $M/2$ jump at Q, option A, fits every condition.
\[ \boxed{\text{Option A}} \]
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