Step 1: Examine the given identity.
The equation provided is:
\[ P e^{x} = Q e^{-x} \]
It is stated that this equality must be valid for every real value of $x$.
Step 2: Simplify the equation.
Multiply both sides by $e^{x}$ to remove the negative exponent:
\[ P e^{2x} = Q \]
Step 3: Use the condition “true for all real x”.
The expression $e^{2x}$ is not constant; it varies continuously with $x$.
Therefore, the only way $P e^{2x}$ can remain equal to the constant $Q$ for all $x$ is if:
\[ P = 0 \]
Substituting $P=0$ back into the original equation gives:
\[ 0 = Q e^{-x} \]
Since $e^{-x}$ is never zero, this equality can hold for all $x$ only when:
\[ Q = 0 \]
Step 4: Verify the options.
(A) $P=Q=0$ satisfies the equation for every real $x$.
(B) $P=Q=1$ does not satisfy the identity for all values of $x$.
(C) $P=1, Q=-1$ clearly fails to satisfy the equation.
(D) $\frac{P}{Q}=0$ alone does not ensure the identity holds for all $x$.
Step 5: Final conclusion.
The only values of $P$ and $Q$ that make the equation true for all real $x$ are:
\[ \boxed{P = Q = 0} \]