Take the dense subset to be those irrationals whose binary expansion $a_m 2^m+\cdots+a_n2^n$ is finite or periodic for the even powers. In other words the binary number formed by the coefficients of the even places is rational.
Then the map from M from $\mathbb Q\times \mathbb P$ to a dense subset of $\mathbb R$ given by interleaving the binary expansions should work. More formally $M(q,p)=T_e(q)+T_o(p)$ where $T_e(r)$ and $T_o(r)$ are defined by taking a binary expansion of $r$ (you can pick the finite one if there is a choice) and mapping $2^n\rightarrow 2^{2n}$ and $2^n\rightarrow 2^{2n+1}$ respectively. i.e.
$T_e(11)=T_e(2^0+2^1+2^3)=2^0+2^2+2^6=69$, $T_o(11)=T_e(2^0+2^1+2^3)=2^1+2^3+2^7=138$.
This map is clearly continuous both ways, 1-1 and maps to a dense subset since any real number has an arbitrarily close rational approximation.
There are of course many possible such examples as you can replace the even integer indices by any fixed infinite subset to place the rational expansion into and this would still work.