Let  n₁ and n₂  be the number of moles of Helium and oxygen gases respectively, and  v₀  be volume of each vessel.  Then from equation of state
$n_1=\frac{P_0{V}_0}{R{T}_0}$  and  $n_2=\frac{4P_0{V}_0}{R\left(2T_0\right)}=\frac{2P_0{V}_0}{R{T}_0}$
$\Rightarrow n_2=2n_1$ 
If T is the final temperature, then the initial total internal energy (kinetic) of the two  gases must be equal to that of the final mixture.
$\begin{array}{l}{\mathrm{U}}_{\mathrm{i}}={\mathrm{n}}_1\left(\frac{3{\mathrm{RT}}_0}{2}\right)+{\mathrm{n}}_2\left(\frac{5\mathrm{R}\left(2{\mathrm{T}}_0\right)}{2}\right)\\ {\mathrm{U}}_{\mathrm{f}}={\mathrm{n}}_1\left(\frac{3\mathrm{RT}}{2}\right)+{\mathrm{n}}_2\left(\frac{5\mathrm{RT}}{2}\right)\\ {\mathrm{U}}_{\mathrm{i}}={\mathrm{U}}_{\mathrm{f}}\text{ gives }\\ \mathrm{T}={\mathrm{T}}_0\left(\frac{3{\mathrm{n}}_1+10{\mathrm{n}}_2}{3{\mathrm{n}}_1+5{\mathrm{n}}_2}\right)={\mathrm{T}}_0\left(\frac{3{\mathrm{n}}_1+20{\mathrm{n}}_1}{3{\mathrm{n}}_1+10{\mathrm{n}}_1}\right)=\frac{23}{13}{\mathrm{T}}_0\end{array}$ 
The final pressure P will be given by
$P\left(2V_0\right)=\left(n_1+n_2\right)RT$

$\Rightarrow P=\frac{\left(n_1+n_2\right)RT}{2V_0}=\frac{3n_{1}RT}{2V_0}$

$\Rightarrow P=\frac{3R}{2V_0}\left(n_1\right)\left(T\right)=\frac{3R}{2V_0}\left(\frac{P_0{V}_0}{R{T}_0}\right)\left(\frac{23}{13}{T}_0\right)=\frac{69}{26}{P}_0$