If you can use the deduction theorem you can assume $p\land s$ and from that follows $p$ and subsequently $q$. Also follows $s$ and subsequently $t$. And since you've proven both $q$ and $t$, $q\land t$ follows. Then you discharge the assumption resulting in the statement to be proved.
$$\begin{align} \tag{1}p\rightarrow q, s\rightarrow t, p\land s &\vdash p\land s&\text{ assumption} \\\tag{2}p\rightarrow q, s\rightarrow t, p\land s &\vdash p&\text{ CE(1)} \\\tag{3}p\rightarrow q, s\rightarrow t, p\land s &\vdash p\rightarrow q&\text{ assumption} \\\tag{4}p\rightarrow q, s\rightarrow t, p\land s &\vdash q&\text{ MP(2,3)} \\\tag{5}p\rightarrow q, s\rightarrow t, p\land s &\vdash s&\text{ CE(1)} \\\tag{6}p\rightarrow q, s\rightarrow t, p\land s &\vdash s\rightarrow t&\text{ assumption} \\\tag{7} p\rightarrow q, s\rightarrow t, p\land s &\vdash t&\text{ MP(5,6)} \\\tag{8} p\rightarrow q, s\rightarrow t, p\land s &\vdash s\land t&\text{ CI(5,7)} \\\tag{9}p\rightarrow q, s\rightarrow t &\vdash (p\land s) \rightarrow (s\land t)&\text{ DT(8)} \\ \end{align}$$
Where CE is conjuction elimination, MP is modus ponens, CI is conjuction introduction and DT is the deduction theorem. Assumption is of course using one of the assumptions to the left of the $\vdash$.
