Determine the angle between the paths of Plane A and Plane B from the airport P.

Plane A is on a bearing of $070^\circ$.
Plane B is on a bearing of $162^\circ$.
The angle $\angle APB$ is the difference between these bearings:

$$\angle APB = 162^\circ - 070^\circ = 92^\circ$$

From the previous part, we know:
Distance PA (from P to A) = $1200 \text{ km}$
Distance PB (from P to B) = $900 \text{ km}$
We use the Law of Cosines in $\triangle APB$ to find the distance AB:

$$AB^2 = PA^2 + PB^2 - 2(PA)(PB)\cos(\angle APB)$$ $$AB^2 = (1200)^2 + (900)^2 - 2(1200)(900)\cos(92^\circ)$$ $$AB^2 = 1,440,000 + 810,000 - 2,160,000\cos(92^\circ)$$ $$AB^2 = 2,250,000 - 2,160,000(-0.034899)$$ $$AB^2 = 2,250,000 + 75,381.84$$ $$AB^2 = 2,325,381.84$$ $$AB = \sqrt{2,325,381.84}$$ $$AB \approx 1524.92 \text{ km}$$

Plane B travels at its original speed, which is $300 \text{ km/h}$.

$$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$$ $$\text{Time} = \frac{1524.92 \text{ km}}{300 \text{ km/h}}$$ $$\text{Time} \approx 5.083067 \text{ hours}$$

The time is 5 full hours.
To find the minutes, multiply the decimal part by 60:

$$0.083067 \text{ hours} \times 60 \text{ minutes/hour} \approx 4.984 \text{ minutes}$$

Rounding to the nearest minute, this is $5 \text{ minutes}$.
So, the time taken is 5 hours and 5 minutes.

The scale is $1 \text{ cm}$ represents $200 \text{ km}$.

$$\text{Length on drawing} = \frac{\text{Actual distance}}{\text{Scale factor}}$$ $$\text{Length on drawing} = \frac{1524.92 \text{ km}}{200 \text{ km/cm}}$$ $$\text{Length on drawing} \approx 7.62 \text{ cm}$$

The time it takes for Plane B to reach Plane A's position is $\boxed{\text{5 hours 5 minutes}}$.
The length of AB on the scale drawing is approximately $\boxed{\text{7.6 cm}}$.