Consider the linear block cipher with block length $n$ and alphabet {0,1}. On the key space of matrices A $\epsilon$ {0,1}$^{(n,n)}$ with $det(A) \cong 1mod(2) $, choose the uniform distribution such that the cryptosystem has perfect secrecy.
The answer I have is:
$P=(0,1)^{n}, K=(0,1)^{(n,n)}$ with $ det(K)\cong 1mod(2)$ where P is the Plaintext and K is the Key.
$E_k(p)=kp=c$ where $c$ is an element of the ciphertext C and $E_k$ is the encryption function.
$Pr(c)=Pr(p \epsilon P)Pr(k \epsilon K)$ where $p$ and $k$ are elements of the plaintext and the key space respectively.
By taking the Uniform distribution of the plaintext as $Pr(p \epsilon P)=\frac{1}{2^n}$ then:
$Pr(c)=\frac{1}{2^n}.Pr(k \epsilon K)$
So $Pr(p|c)=\frac{Pr(p \cap c)}{Pr(c)}=\frac{Pr(k)}{Pr(c)}=\frac{Pr(k)}{Pr(k).\frac{1}{2^n}}=\frac{1}{2^n}$=$Pr(p)$.
Is the above answer satisfactory?
I know that in order for a cryptosystem to have perfect secrecy $Pr(p|c)=Pr(p)$ and so $Pr(p)$ and $Pr(c)$ are independent for all plaintexts p and all ciphertexts c.
So in this case $Pr(p|c)=Pr(p)=\frac{1}{2^n}$.
I also have two queries in the answer I found online given above. These are
1) $\frac{Pr(k)}{Pr(k).\frac{1}{2^n}}=\frac{1}{2^n}$ where surely this would in fact would simplify to $2^n$ instead of $\frac{1}{2^n}$.
2) $Pr(p|c)=\frac{Pr(p \cap c)}{Pr(c)}=\frac{Pr(k)}{Pr(c)}$ where I'm unsure how $Pr(p \cap c)=Pr(k)$
Edit: I have just found a formula for $Pr(p|c)$, namely
$Pr(p|c)$=$\frac{Pr(c|p)Pr(p)}{Pr(c)}$=$\frac{Pr(k(p))Pr(p)}{Pr(c)}$=$\frac{Pr(c)Pr(p)}{Pr(c)}$=$\frac{Pr(c)\frac{1}{2^n}}{Pr(c)}$=$\frac{1}{2^n}$.
So using this instead of $Pr(p|c)=\frac{Pr(p \cap c)}{Pr(c)}=\frac{Pr(k)}{Pr(c)}=\frac{Pr(k)}{Pr(k).\frac{1}{2^n}}=\frac{1}{2^n}$=$Pr(p)$.
My answer is now:
$P=(0,1)^{n}, K=(0,1)^{(n,n)}$ with $ det(K)\cong 1mod(2)$ where P is the Plaintext and K is the Key.
$E_k(p)=kp=c$ where $c$ is an element of the ciphertext C and $E_k$ is the encryption function.
$Pr(c)=Pr(p \epsilon P)Pr(k \epsilon K)$ where $p$ and $k$ are elements of the plaintext and the key space respectively.
By taking the Uniform distribution of the plaintext as $Pr(p \epsilon P)=\frac{1}{2^n}$ then:
$Pr(c)=\frac{1}{2^n}.Pr(k \epsilon K)$
$Pr(p|c)$=$\frac{Pr(c|p)Pr(p)}{Pr(c)}$=$\frac{Pr(k(p))Pr(p)}{Pr(c)}$=$\frac{Pr(c)Pr(p)}{Pr(c)}$=$\frac{Pr(c)\frac{1}{2^n}}{Pr(c)}$=$\frac{1}{2^n}$=$Pr(p)$.