1. Balance condition in meter bridge

At null deflection, the ratio of the resistances in the two gaps equals the ratio of resistances of the wire segments:

$\frac{R_{\text{left gap}}}{R_{\text{right gap}}}=\frac{R_{AP}}{R_{PB}}$

Since the two resistances connected in the gaps are equal, we have

$R_{\text{left gap}}=R_{\text{right gap}}$

Hence, at balance:

$R_{AP}=R_{PB}$

So point $P$ divides the wire into two parts of equal resistance, not necessarily equal length because the wire is non-uniform.

2. Given variation of resistance per unit length

We are told:

$\frac{dR}{d\ell} \propto \frac{1}{\sqrt{\ell}}$

Let

$\frac{dR}{d\ell}=\frac{k}{\sqrt{\ell}}$

where $k$ is a constant.

Then the resistance from end $A$ up to a point at distance $x$ is

$R(x)=\int_0^x \frac{k}{\sqrt{\ell}}\, d\ell$

$R(x)=k\int_0^x \ell^{-1/2} d\ell$

$R(x)=k\left[2\sqrt{\ell}\right]_0^x=2k\sqrt{x}$

So,

$R_{AP}=2k\sqrt{x}$

where $x=AP$.

3. Resistance of the remaining part $PB$

Total wire length is $1\text{ m}$, so total resistance is

$R_{AB}=2k\sqrt{1}=2k$

Thus resistance of segment $PB$ is

$R_{PB}=R_{AB}-R_{AP}=2k-2k\sqrt{x}=2k(1-\sqrt{x})$

4. Apply balance condition

Since $R_{AP}=R_{PB}$,

$2k\sqrt{x}=2k(1-\sqrt{x})$

Cancel $2k$:

$\sqrt{x}=1-\sqrt{x}$

$2\sqrt{x}=1$

$\sqrt{x}=\frac{1}{2}$

$x=\frac{1}{4}=0.25\text{ m}$

5. Check options

The correct option is:

$\boxed{0.25\text{ m}}$

So, Option B is correct.