If p, q, r, s are statements, where, $p: A^2-B^2=(A-B)(A+B) ; A, B$ are matrices, $A B \neq B A$ q: $5 \leqslant 5$ r: ${ }^8 \mathrm{C}_1+{ }^8 \mathrm{C}_2+{ }^8 \mathrm{C}_3+\ldots \ldots \ldots . .+{ }^8 \mathrm{C}_8=256$ s: Maximum value of ${ }^8 \mathrm{C}_{\mathrm{r}}$ is 70 then the statement from the following having truth value true is ....

Statement p: $A^2-B^2=(A-B)(A+B)$ with matrices $A,B$ where $AB \neq BA$.

Expanding the right side gives $(A-B)(A+B) = A^2 + AB - BA - B^2$. For $A^2-B^2$ to equal this expression, $AB$ must equal $BA$, but by hypothesis $AB \neq BA$. Thus, $p$ is false.

Statement q: $5 \leqslant 5$ is true since equality holds.

Statement r: ${}^8C_1 + {}^8C_2 + \ldots + {}^8C_8$.

The full sum from $k=0$ to $8$ is $2^8 = 256$, so the given sum is $256 - {}^8C_0 = 255$. Since $255 \neq 256$, $r$ is false.

Statement s: Maximum value of ${}^8C_r$ occurs at $r=4$ and equals $\frac{8!}{4!4!} = 70$, so $s$ is true.

Evaluating options:

Option A: $(p \wedge \sim r) \vee (\sim q \wedge \sim s) = (F \wedge T) \vee (F \wedge F) = F \vee F = F$

Option B: $(p \vee \sim q) \leftrightarrow (\sim r \rightarrow s) = (F \vee F) \leftrightarrow (T \rightarrow T) = F \leftrightarrow T = F$

Option C: $(p \leftrightarrow q) \wedge (\sim p \vee \sim q) = (F \leftrightarrow T) \wedge (T \vee F) = F \wedge T = F$

Option D: $(s \vee \sim p) \leftrightarrow (\sim p \wedge \sim r) = (T \vee T) \leftrightarrow (T \wedge T) = T \leftrightarrow T = T$

The only true statement is Option D.