Solution:

The answer is 6. Let us choose a Cartesian reference frame placing vertex $B$ at the origin and the $x$ and $y$ axes respectively along segments $AB$ and $BC$.

Consider the stretch that Nicola covers from $T$, the midpoint of $AD$, to vertex $A$ (see figure 1). Since Nicola and Michela run at the same speed, when Nicola is at $N$ and Michela is at $M$, it follows that $TN = BM$. By central symmetry with respect to the point $P$, with coordinates $\left(\frac{\sqrt{18}}{2}, \frac{\sqrt{18}}{4}\right)$, it follows that during this stretch the dog Pallino stays fixed at point $P$.

Now consider the stretch that Nicola covers from $A$ to the midpoint of $AB$ (see figure 2). Again by the fact that Nicola and Michela run at the same speed, we have $BN + BM = 3 \frac{\sqrt{18}}{2}$, so denoting by $x$ and $y$ the coordinates of the midpoint of $MN$, we obtain the relation:
$$x + y = \frac{3}{4} \sqrt{18}$$
This relation implies that during this stretch Pallino moves along a line, in particular it covers the segment $PQ$ where $Q$ is the point with coordinates $\left(\frac{\sqrt{18}}{4}, \frac{\sqrt{18}}{2}\right)$ at which Pallino is located when Michela is at $C$ and Nicola is at the midpoint of $AB$.

Now the situation is identical to the initial one, rotated by an angle of $90^\circ$. Pallino therefore covers (see figure 3) the square $PQRS$, stopping for a certain time at the vertices. This square has side equal to:
$$\sqrt{\left(x_{P}-x_{Q}\right)^{2}+\left(y_{P}-y_{Q}\right)^{2}}=\sqrt{\left(\frac{\sqrt{18}}{4}-\frac{\sqrt{18}}{2}\right)^{2}+\left(\frac{\sqrt{18}}{2}-\frac{\sqrt{18}}{4}\right)^{2}}=\frac{3}{2}.$$
Therefore, when Nicola has completed the lap of the park, Pallino has covered $4 \cdot \frac{3}{2} = 6$ kilometers.