Net magnetic field at $P$ is

$B=\frac{\mu_0 I_1}{2 \pi x \times 10^{-2}}-\frac{\mu_0 I_2}{2 \pi(4-x) \times 10^{-2}}$

$=\frac{\mu_0}{2 \pi \times 10^{-2}}\left(\frac{I_1}{x}+\frac{I_2}{x-4}\right)$

For $B$ to be minimum,

$\frac{d B}{d x}=0 \Rightarrow \frac{d}{d x}\left(\frac{I_1}{x}+\frac{I_2}{x-4}\right)=0$

$\Rightarrow-\frac{I_1}{x^2}-\frac{I_2}{(x-4)^2}=0 \Rightarrow-\frac{I_1}{x^2}=\frac{I_2}{(x-4)^2}$

$\Rightarrow \frac{I_1}{I_2}=-\left(\frac{x}{x-4}\right)^2$

with $x=1 \,cm$, we have

$\frac{I_1}{I_2}=-\left(\frac{1}{1-4}\right)^2=-\frac{1}{9} \text { or } \frac{I_2}{I_1}=-9$

Here, negative sign shows that they are anti-parallel.