I'm having a bit of trouble understanding this question:
Consider an isolated system made up of two Einstein solids A and B each containing 2 particles and separated by a diathermal wall. The overall sample space is
$$Ω(E)=\sum_{E(A)}^{\_} Ω_A(E_A) × Ω_B(E − E_A)$$
and the probability of system A having energy $E_A$ is $$P(E_A) = \frac{|Ω_A(E_A)||Ω_B(E − E_A)|}{|Ω(E)| }$$
Complete the following table and sketch $P(E_A)$ against $E_A$.
I've filled in the table and I believe it looks as follows:
Now the problem I have is finding out what are the actual calculations you do to find the $Ω_A(E_A)$ and $Ω_A(6-E_A)$ values.
From the wikipage on Einstein solids, we get the formula for $\Omega$ to be
$$\Omega = \frac{(q+N'-1)!}{q!(N'-1)!}$$
where $q$ is the amount of energy quanta that you can distribute over the $N'$ oscillators. Since you don't give enough context, I don't know what $N'$ is for sure. But it seems that if we pick $q=E_A$ and $N'=2$ we get
$$\Omega = \frac{(E_A+1)!}{E_A!(1)!}=E_A+1$$
which corresponds exactly with the numbers you get. This would mean that your Einstein solids are 1D, i.e. there is 1 degree of freedom for each particle.