Position of Point in Argand Plane After Two Displacements

To find the position of the point $P$ in the Argand plane, we represent the two displacements as complex numbers in vector form.

Let the East direction correspond to the positive real axis (positive $x$-axis) and the North direction correspond to the positive imaginary axis (positive $y$-axis).

1. First Displacement:

   • The man walks a distance of $3$ units towards the North-East ($N\ 45^{\circ}\ E$) direction from the origin.
   • The angle made by this direction with the positive real axis is:
     $\theta_1 = 45^{\circ} = \frac{\pi}{4}$
   • Therefore, the first displacement vector $z_1$ is given by:
     $z_1 = 3 e^{i\pi/4}$

2. Second Displacement:

   • From the new position, he walks a distance of $4$ units towards the North-West ($N\ 45^{\circ}\ W$) direction.
   • The North-West direction makes an angle of $45^{\circ}$ to the west of North, which corresponds to an angle with the positive real axis of:
     $\theta_2 = 90^{\circ} + 45^{\circ} = 135^{\circ} = \frac{3\pi}{4}$
   • Therefore, the second displacement vector $z_2$ is:
     $z_2 = 4 e^{i 3\pi/4}$

3. Total Displacement / Position of $P$:
   The position of point $P$ is the vector sum of both displacements:
   $z = z_1 + z_2 = 3 e^{i\pi/4} + 4 e^{i 3\pi/4}$

   We can rewrite $e^{i 3\pi/4}$ using the properties of complex exponentials:
   $e^{i 3\pi/4} = e^{i(\pi/4 + \pi/2)} = e^{i\pi/4} \cdot e^{i\pi/2}$
   Since $e^{i\pi/2} = \cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right) = i$, we have:
   $e^{i 3\pi/4} = i e^{i\pi/4}$

   Substituting this back into the expression for $z$:
   $z = 3 e^{i\pi/4} + 4(i e^{i\pi/4}) = (3 + 4i)e^{i\pi/4}$

Thus, the position of $P$ in the Argand plane is $(3 + 4i)e^{i\pi/4}$.

Therefore, the correct option is (D).