The equivalent resistance between P and Q is:

8 Ω

The correct answer is Option (1) → 8 Ω

Given: We need to find the equivalent resistance between points $P$ and $Q$.

From the circuit, observe that the $3\ \Omega$ resistor (vertical) and the $6\ \Omega$ resistor (left side) are connected between the same two points — both joining the top point $P$ to the central node.

Hence, they are in parallel:

$R_{PO} = \left( \frac{1}{3} + \frac{1}{6} \right)^{-1} = 2\ \Omega$

Now, the circuit reduces to a triangle between points $P$, $O$, and $Q$ with the following resistances:

• $P$–$O$ branch: $2\ \Omega$
• $O$–$Q$ branch: $10\ \Omega$
• $P$–$Q$ branch (outer arc): $24\ \Omega$

The path through $O$ gives a total resistance of:

$R_{POQ} = R_{PO} + R_{OQ} = 2 + 10 = 12\ \Omega$

This $12\ \Omega$ path is in parallel with the direct $24\ \Omega$ branch between $P$ and $Q$:

$R_{PQ} = \left( \frac{1}{24} + \frac{1}{12} \right)^{-1}$

Calculate:

$R_{PQ} = \frac{24 \times 12}{24 + 12} = \frac{288}{36} = 8\ \Omega$

Hence, the equivalent resistance between P and Q is: $R_{PQ} = 8\ \Omega$