Parallel Connection of Capacitor Assume Two Capacitor filled with different dielectrics with dielectric constant $K_1$ and $K_2$ respectively in Parallel connection.  Then we have same voltage $V$ on both capacitors. Therefore, $$\dfrac{E}{K_1}d=\dfrac{E}{K_2}d.$$ So, $K_1= K_2$. This means we have always two dielectrics with same dielectric constant. This is clear contradiction, Where I am mistaken?
 A: You seems to assume both capacitors has the same plate separation $d$. So, lets assume that. Assume there is no dielectric material. Therefore, nicely $Ed = Ed$ in both capacitors. Which is nice. :).
Now, I think I understand your confusion. Have an isolated capacitor with electric field inside plates of $E$. Insert dielectric $K$. Under this case, the electric field now falls $E/K$. But under this scenario, there were also a voltage drop. Why? Because here the charge is conserved because capacitors are isolated.
If you force potential to be the same, then charge increase will contribute on electric field with $K$, and electric field will fall $K$, canceling out, then $E_1 = E_2$.
Since $Q = CV$, the electric field after dielectric:
$$
E_{after} = 
\frac{\sigma}{\epsilon} = 
\frac{Q_{after}}{\epsilon_0 KA} = 
\frac{C_{after}V}{\epsilon_0 KA} = 
\frac{KC_{before}V}{\epsilon_0 KA} = 
\frac{C_{before}V}{\epsilon_0 A} = 
E_{before}
$$
Then, decrease by $K$ of electric field after dielectric is inserted field under condition of constant $V$ does not happen. Anyway, the neat answer of @AnubhavGoel address this far simpler than mine.
A: $E_1 = E_2$ . since $E$ is independent of dielectric as long as potential b/w plates is constant.
$$E= =  -\frac{dV}{dr}$$
So, it is independent of dielectric b/w it.
So, correct statement would be 
$$E_1d = E_2d$$
$$Ed = Ed$$
