Two identical inductors are connected in two different...

Let the inductance of each identical inductor be $L$.

For configuration (P):
The two inductors are connected in series. The points $a$ and $b$ are located across the first inductor only. The current flowing through this inductor is $I(t)$.
The induced emf between points $a$ and $b$ is given by:
$E_P = L \dfrac{dI}{dt}$

For configuration (Q):
The two inductors are connected in parallel. The points $a$ and $b$ are the common nodes of this parallel combination.
The equivalent inductance $L_{eq}$ between $a$ and $b$ is:
$L_{eq} = \dfrac{L \times L}{L + L} = \dfrac{L}{2}$
The total current entering the parallel combination is $I(t)$. The induced emf between points $a$ and $b$ is:
$E_Q = L_{eq} \dfrac{dI}{dt} = \dfrac{L}{2} \dfrac{dI}{dt}$

Now, finding the ratio of the induced emfs:
$\dfrac{E_P}{E_Q} = \dfrac{L \dfrac{dI}{dt}}{\dfrac{L}{2} \dfrac{dI}{dt}} = 2$

Answer: $2$