For a positively charged particle moving in a $x - y$ plane initially along the $x$-axis, there is a sudden change in its path due to the presence of electric and/or magnetic fields beyond $P$. The curved path is shown in the $x - y$ plane and is found to be non-circular. Which one of the following combinations is possible 

Initially the particle moves along $+x$, so $\vec{v}=v\hat{i}$. For the motion to remain in the $x-y$ plane, the net force should have no $\hat{k}$ component. Check each option using $$\vec{F}=q(\vec{E}+\vec{v}\times\vec{B}).$$ For option $A$, $$\vec{B}=b\hat{i}+c\hat{k}.$$ Then $$\vec{v}\times\vec{B}=v\hat{i}\times(b\hat{i}+c\hat{k})=-vc\hat{j}.$$ So the force is only along $-y$. With only magnetic force, speed stays constant and the path is circular, not non-circular. So $A$ is not possible. For option $B$, $$\vec{E}=a\hat{i},\quad \vec{B}=c\hat{k}+a\hat{i}.$$ Then $$\vec{v}\times\vec{B}=v\hat{i}\times(c\hat{k}+a\hat{i})=-vc\hat{j}.$$ Hence there is an electric force along $+x$ and a magnetic force along $-y$. The $x$-component changes the speed, while the $y$-component bends the path. Therefore the trajectory in the $x-y$ plane can be non-circular. So this is possible. For option $C$, $$\vec{B}=c\hat{j}+b\hat{k}.$$ Then $$\vec{v}\times\vec{B}=v\hat{i}\times(c\hat{j}+b\hat{k})=vc\hat{k}-vb\hat{j}.$$ There is a $\hat{k}$ component, so motion will leave the $x-y$ plane. Not possible. For option $D$, $$\vec{E}=a\hat{i},\quad \vec{B}=c\hat{k}+b\hat{j}.$$ Then $$\vec{v}\times\vec{B}=v\hat{i}\times(c\hat{k}+b\hat{j})=-vc\hat{j}+vb\hat{k}.$$ Again there is a $\hat{k}$ component, so motion will not stay in the $x-y$ plane. Not possible. After comparing all options, only $B$ gives motion confined to the $x-y$ plane and non-circular trajectory.