If there is nontrivial isotrivial elliptic fibration, what will be type of singular fiber(since we have the kodaira's table of singular fiber). Is it a contradiction since in kodaira's table, smooth fiber around singular fiber seems non isomorphic. I am a bit confused?
I think one ususally uses isotrivial in this case to mean that the smooth fibers are all isomorphic. Then it is certainly the case that there can be singular fibers.
Which singular fibers can appear depends on the $j$ invariant. The $j$ invariants at which they can appear are the same as in a non-isotrivial fibration:
$I_0^*$ - any $j$ invariant.
$III, III^*$ - only $j=1728$.
$II, II^*, IV, IV^*$ - only $j=0$.
It is easy to construct examples of these:
For $j=1728$, the family $y^2= x^3 - t^k x$ will be of type $III$ if $k=1$, $I_0^*$ if $k=2$, and $III^*$ if $k=3$.
For $j=0$, the family $y^2 = x^3 - t^k x$ will be of type $(II,IV,I_0^*, IV^*, II^*)$ respectively as $k=(1,2,3,4,5)$.
And $y^2 =x^3 - t^2 ax -t^3 b$ will always have type $I_0^*$ when $y^2=x^3-ax-b$ is smooth.
I think you can check all these by Tate's algorithm, for instance.
I think in Kodaira's table he gives an example of a family for each type of singular fiber. But these are not the only examples.