Interpret the circuit

The two batteries with internal resistances are connected in parallel between points $P_1$ and $P_2$, and the $10\Omega$ resistor is also connected across the same two points.

Let the potential difference across the resistor be $V = V_{P_1}-V_{P_2}.$

Then current through the $10\Omega$ resistor is $I_R = \frac{V}{10}$ taken positive from $P_1$ to $P_2$.

Write branch currents

For the $5\text{ V}$ battery with internal resistance $2\Omega$: assuming its positive terminal is at $P_1$ (as usual in such figure), current from $P_1$ to $P_2$ in that branch is $I_1 = \frac{V-5}{2}.$

For the $2\text{ V}$ battery with internal resistance $1\Omega$: $I_2 = V-2.$

Through the resistor: $I_R = \frac{V}{10}.$

Apply KCL at node $P_1$

Since there is no external connection, algebraic sum of currents from $P_1$ to $P_2$ through all three branches must be zero: $\frac{V-5}{2} + (V-2) + \frac{V}{10} = 0.$

Solve for $V$

Multiply by $10$: $5(V-5) + 10(V-2) + V = 0$ $5V - 25 + 10V - 20 + V = 0$ $16V - 45 = 0$ $V = \frac{45}{16} = 2.8125\text{ V}.$

Find current in the $10\Omega$ resistor

$I_R = \frac{V}{10} = \frac{2.8125}{10} = 0.28125\text{ A}.$

So the current is approximately $0.28\text{ A}$ flowing from $P_1$ to $P_2$.

Match with the given options

Closest option is: $\boxed{0.27\text{ A from } P_1 \text{ to } P_2}$

Hence, the correct option is D.

Comparison with stored answer

Stored correct answer is C: $0.03\text{ A } P_2 \to P_1$.

This does not match the circuit analysis. The likely correct answer is D.