A heavy nucleus N, at rest, undergoes fission $$N \rightarrow P + Q$$, where P and Q are two lighter nuclei. Let $$\delta = M_N − M_P − M_Q$$, where $$M_P$$, $$M_Q$$ and $$M_N$$ are the masses of P, Q and N, respectively. EP and EQ are the kinetic energies of P and Q, respectively. The speeds of P and Q are $$v_P$$ and $$v_Q$$, respectively. If c is the speed of light, which of the following statement(s) is(are) correct?

The parent nucleus $$N$$ is initially at rest, so its four-momentum is $$\left(M_Nc,\,0\right)$$.

After fission we have two product nuclei $$P$$ and $$Q$$ with four-momenta

$$P_P=\left(E_P^t/c,\;\vec p_P\right),\qquad P_Q=\left(E_Q^t/c,\;\vec p_Q\right)$$
where $$E_P^t=E_P+M_Pc^2,\;E_Q^t=E_Q+M_Qc^2$$ are the total (rest + kinetic) energies and the kinetic energies are $$E_P,\;E_Q$$.

Step 1 : Momentum conservation
Because the initial momentum is zero, the two fragments recoil in opposite directions with equal magnitude:

$$\vec p_P+\vec p_Q=0\;\;\Longrightarrow\;\;p_P=p_Q\;(\equiv p).$$

Step 2 : Energy conservation
The time component of four-momentum gives

$$E_P^t+E_Q^t=M_Nc^2.$$

Rewrite in terms of kinetic energies:

$$\bigl(E_P+M_Pc^2\bigr)+\bigl(E_Q+M_Qc^2\bigr)=M_Nc^2.$$

Collecting the rest-mass terms,

$$E_P+E_Q=\bigl(M_N-M_P-M_Q\bigr)c^2.$$

By definition $$\delta = M_N-M_P-M_Q$$, therefore

$$E_P+E_Q=c^2\delta\;.\qquad-(1)$$

Equation $$(1)$$ is independent of whether the fragments move non-relativistically or relativistically; it follows only from energy-momentum conservation. Hence

Option A is always correct.

Why the other options are not guaranteed

Option B claims $$E_P=\dfrac{M_P}{M_P+M_Q}\,c^2\delta$$. To decide this we need the individual kinetic energies. Using only conservation laws we have two unknowns ($$E_P,\,E_Q$$) but just one equation $$(1)$$, so their individual values are not fixed by masses alone. The quoted expression can be obtained only under the extra, non-relativistic assumption $$E\ll M c^2$$, which is not stated in the question. Therefore Option B is not necessarily true.

Option C states $$v_P/v_Q = M_Q/M_P$$. Relativistically $$v=\dfrac{pc^2}{E^t}$$, so

$$\frac{v_P}{v_Q}=\frac{E_Q^t}{E_P^t}\neq\frac{M_Q}{M_P}$$ in general. Again, the quoted ratio holds only in the non-relativistic limit $$p\ll Mc$$ where $$E^t\approx Mc^2$$. Hence Option C is not generally valid.

Option D proposes $$p=c\sqrt{2\mu\delta}\;,\;\; \mu=\dfrac{M_PM_Q}{M_P+M_Q}$$. Starting from the exact relation

$$E_P^t=\sqrt{(pc)^2+M_P^2c^4},\;E_Q^t=\sqrt{(pc)^2+M_Q^2c^4},$$

substituting into energy conservation gives a quadratic equation for $$p$$ that does not simplify to the quoted form. The expression $$p=c\sqrt{2\mu\delta}$$ again appears only when the kinetic energies are assumed much smaller than the rest energies (so $$E_P\approx p^2/2M_P$$ etc.). Without that approximation Option D is incorrect.

Hence, under the general, relativistically correct treatment of nuclear fission, only

Option A which is: $$E_P+E_Q=c^2\delta$$

is universally correct.