Out of 52 cards, 4 cards can be drawn in $C(52,4)$ ways. There are 13 diamonds cards out of which 3 cards can be drawn in $C(13,3)$ ways.
Also one spade card out of 13 spade cards can be drawn in $C(13,1)$ ways. So, favourable number of cases $=C(13,3) \times \mathrm{C}(13,1)$
Hence, required probability
$\begin{aligned}
&=\frac{C(13,3) \times C(13,1)}{C(52,4)}=\frac{\frac{13 \times 12 \times 11}{3 \times 2 \times 1} \times 13}{\frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1}} \\
&=\frac{13 \times 12 \times 11 \times 13 \times 4}{52 \times 51 \times 50 \times 49}=\frac{286}{20825}
\end{aligned}$