Since, $F_{ext}=0$

So,

According to conservation of linear momentum,

${\left(p_{\mathrm{total} }\right)}_{\mathrm{initial}}={\left(p_{\mathrm{total} }\right)}_{\mathrm{final}}=0$

That's why momentum of both lighter nuclei will be equal and opposite to each other as we can see in above diagram.

Now here some energy released due to mass defect which can be written as,

Energy released $=\left(\mathrm{\Delta }m\right){\mathrm{c}}^2=\delta {\mathrm{c}}^2$

Where, $\mathrm{\Delta }m=\delta =$mass defect

and, $\delta =M_N-M_P-M_Q$

Now this energy released due to mass defect convert in kinetic energy of daughter nuclei so,

Energy released$=\left(\mathrm{\Delta }m\right){\mathrm{c}}^2={\mathrm{c}}^2\mathrm{\delta }=K{E}_P+K{E}_Q$

Where,

$$\begin{gathered} K{E}_P=\frac{p^2}{2M_P}, K{E}_Q=\frac{p^2}{2M_Q} \\ \Rightarrow K{E}_p:K{E}_Q=\frac{1}{{ M}_P}:\frac{1}{{ M}_Q}=M_Q:M_P \end{gathered}$$

$K{E}_P=\left(\frac{M_Q}{M_p+M_Q}\right)\delta c^2$

$K{E}_Q=\left(\frac{M_p}{M_P+M_Q}\right)\mathrm{\delta }{c}^2$

$K{E}_P+K{E}_Q=\delta c^2$

$\frac{p^2}{2{\mathrm{M}}_P}+\frac{p^2}{2{\mathrm{M}}_Q}=\delta c^2$

$\Rightarrow p=c\sqrt{2\left(\frac{M_P{M}_Q}{M_P+M_Q}\right)\delta }$

Hence, options $\left(1\right),\left(3\right),\left(4\right)$are correct.