Each battery shown in the figure has an emf equal to 5 V ....

Let the potential at the bottom-middle junction be $0\text{ V}$ and at the top-middle junction be $V$.

From the polarities of the batteries, the potential at the bottom-left node is $5\text{ V}$ and at the bottom-right node is $5\text{ V}$.

Similarly, the potential at the top-left node is $V + 5\text{ V}$ and at the top-right node is $V + 5\text{ V}$.

The potential difference across the left resistor is $(V + 5) - 5 = V$.

The potential difference across the middle resistor is $V - 0 = V$.

The potential difference across the right resistor is $(V + 5) - 5 = V$.

Applying Kirchhoff's Current Law at the top nodes, the total current leaving through the resistors must be zero:

$\dfrac{V}{R_1} + \dfrac{V}{R_2} + \dfrac{V}{R_3} = 0$

$V \left( \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} \right) = 0$

Since the resistances are positive, $V = 0$.

As the potential difference across each resistor is zero, no current flows in any branch of the circuit.

Since the current is zero everywhere, the magnetic field $B$ produced by the circuit at point $P$ is $0$.

Answer: $0$