In a vernier caliper, N divisions of the vernier scale coincide...

First, we determine the least count (L.C.) of the vernier caliper.
Given $1 \text{ MSD} = a$ and $N \text{ VSD} = (N-1) \text{ MSD}$.
$\Rightarrow 1 \text{ VSD} = \dfrac{N-1}{N} a$
$\text{L.C.} = 1 \text{ MSD} - 1 \text{ VSD} = a - \dfrac{N-1}{N} a = \dfrac{a}{N}$

When the jaws are in contact, the zero of the vernier scale is to the right of the main scale zero, which means the zero error is positive.
Zero error $= + k \times \text{L.C.} = + \dfrac{ka}{N}$

During the measurement, the main scale reading is $M$ divisions, which equals $Ma$.
The vernier scale reading is $p \times \text{L.C.} = \dfrac{pa}{N}$.
Measured value $= Ma + \dfrac{pa}{N}$

The true length of the rod is obtained by subtracting the zero error from the measured value:
True length $= \text{Measured value} - \text{Zero error}$
True length $= \left( Ma + \dfrac{pa}{N} \right) - \dfrac{ka}{N}$
True length $= a \left[ M + \dfrac{p-k}{N} \right]$

Answer: a \left[ M + \dfrac{p-k}{N} \right]