A man walks 5 km due north and then 4 km due west. What is his bearing from his starting point?

Let point A be the origin.
The man walks $5 \text{ km}$ due north from A to an intermediate point, let's call it B. So, the coordinates of B relative to A are $(0, 5)$.
From B, he walks $4 \text{ km}$ due west to point C. So, the coordinates of C relative to B are $(-4, 0)$.
Therefore, the coordinates of C relative to A are $(-4, 5)$.
This movement forms a right-angled triangle $\triangle ABC$, with the right angle at B.
The length of side AB is $5 \text{ km}$ (north).
The length of side BC is $4 \text{ km}$ (west).

We need to find the angle $\angle BAC$, which is the angle between the North line (AB) and the line segment AC.
In the right-angled triangle $\triangle ABC$:

Point C is located North-West of point A. The angle $\angle BAC$ is the angle measured from the North line (AB) towards the West.
Bearings are measured clockwise from the North line ($0^\circ$).
For a point in the North-West quadrant, the bearing is $360^\circ$ minus the angle measured from the North line towards the West.

Rounding to one decimal place, the bearing is $321.3^\circ$.