A

Step 1: Calculate the total resistance in the circuit before connecting the ammeter.
The resistors are connected in series:
R_total = R₁ + R₂ = 400 Ω + 800 Ω = 1200 Ω.

Step 2: Determine the current in the circuit without the ammeter.
Using Ohm's Law:
I = \frac{V}{R} = \frac{6 \text{ V}}{1200 \text{ Ω}} = 0.005 ext{ A} = 5 ext{ mA}.

Step 3: Consider the ammeter in series. The ammeter has a resistance of 10 Ω.
The new total resistance:
R_total = R₁ + R₂ + R_ammeter = 1200 Ω + 10 Ω = 1210 Ω.

Step 4: Calculate the current with the ammeter connected.
I' = \frac{V}{R'} = \frac{6 ext{ V}}{1210 ext{ Ω}} \approx 0.004964 ext{ A} = 4.964 ext{ mA}.

Step 5: Now, for the voltmeter, its resistance is 1000 Ω connected across the 400 Ω resistor, reducing the total resistance in that segment.
The equivalent resistance for the 400 Ω resistor (R₁) and the voltmeter (R_v) in parallel:
R_parallel = \frac{R₁ \cdot R_v}{R₁ + R_v} = \frac{400 \cdot 1000}{400 + 1000} = \frac{400000}{1400} \approx 285.71 Ω.

Step 6: The equivalent circuit with resistors in series gives:
R_total = 285.71 Ω + 800 Ω + 10 Ω = 1095.71 Ω.
The voltage across the 400 Ω resistor using the current with the ammeter:
V_R1 = I' \cdot R₁ = 4.964 ext{ mA} \cdot 400 Ω \approx 1.9856 ext{ V} \approx 1985.6 ext{ mV}.

Step 7: For values N and P, based on the earlier equations:
N = 1985.6, and P = 19 (from the voltmeter reading). Thus,
N = 1210 and P = 19 (the values for the equation given).
Therefore, N = 1210 and P = 19.