Homomorphic encryption based on XOR Consider the values $a$ and $b$ are encrypted as $c_1=(a \oplus D)$ and $c_2=(b \oplus D)$.

My question is: can we derive $b+a$ from any combination of $c_1$ and $c_2$? 

Note: 


I admit that the $D$ should have not been used for both $a$ and $b$. For now assume we can use it.
Basically I define the values ($a$ and $b$) in $Z_p$, accordingly the result in $Z_p$.
Based on the additional details in the comments, it seems like your question is: given $c_1=a\oplus d$ and $c_2=b\oplus d$, can we get $(a+b)\oplus d$. Where $a,b,d\in\mathbb{Z}_p$, $+$ is addition modulo $p$, and $\oplus$ is a bitwise XOR of the values, then taken modulo $p$.

Or put another way, is there an operation $\boxplus$, such that $(a\oplus d)\boxplus (b\oplus d)\equiv (a+b)\oplus d\bmod{p}$?

Hopefully I've understood the problem correctly.

The answer is no, except in the trivial case of $p=2$, or more generally (if $p$ is not necessarily a prime) when $p=2^q$ for some prime $q$. That is because in these two cases, in the finite field $\mathbb{Z}_p$, $\oplus$ and $+$ are traditionally the same operation (note that by $+$ I am not talking about integer addition modulo $p$, I am referring to addition in the finite field). Or put another way $+$ is defined as bitwise xor in these fields.

In the case where $p\neq 2$ and $p\neq 2^q$ for some prime $q$, the bitwise XOR destroys all algebraic structure necessary for there to be an operation $\boxplus$. Now, what you can do is use $+$ instead of $\oplus$. Then you have $(a+d) + (b+d) = (a+b) + (d+d)$ and to decrypt, subtract $(d+d)$. If doing this, you should use different $d$ values for each message. In other words $(a+d_1) + (b+d_2) = (a+b) + (d_1+d_2)$ and to decrypt, subtract $(d_1+d_2)$. The $d_i$ values must be randomly chosen.

Full disclosure: My answer is based on intuition. It might be hard to prove that $\boxplus$ does not exist, but it seems very unlikely that it does. Basically we need associativity between $\oplus$ and $\boxplus$ (i.e., $(a\oplus b)\boxplus c = a \oplus (b\boxplus c)$ and $\boxplus$ must be commutative ($a\boxplus b = b\boxplus a$). There are lots of candidates for commutativity, but I can't think of any for that type of associativity.