With $X\sim\operatorname{bin}(3,1/2)$, $Y\sim\operatorname{unif}\{1,2,3,4\}$ being statistically independent we are asked to find $P(X=Y)$
This is how I tried solving the problem:
\begin{align*}
P(X=Y) &= \sum_{k=1}^{3} P(X=k,Y=k) \\
&= \sum_{k=1}^3 P(X=k)P(Y=k) \\
&= \sum_{k=1}^3 \binom{3}{k} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{3-k} \frac{1}{4}
\end{align*}
but the answer key doesn't have $\left(\frac{1}{2} \right)^{3-k}$ and instead:
\begin{align*}
&= \sum_{k=1}^3 \binom{3}{k} \left(\frac{1}{2}\right)^3 \frac{1}{4} \\
&=\frac{1}{32} \left[ \binom{3}{1} + \binom{3}{2} + \binom{3}{3} \right] \\
&=\frac{7}{32} \
\end{align*}
Why is $\left(\frac{1}{2}\right)^{3-k}$ left out even though for binomial distribution: $P(n=k)=\binom{n}{k} p^n q^{n-k}$?
It looks like you just made a simple mistake when you were writing down the probability density function of the binomial distribution. From the second to third line, you should instead have
$$\sum_{k=1}^{3} P(X = k)P(Y = k) = \sum_{k=1}^{3} {3\choose k}\left(\frac{1}{2}\right)^k \left(\frac{1}{2}\right)^{3 - k}\frac{1}{4}$$
This will simplify into your textbook's answer.