Calculating entropies of attributes
Consider the formula for Entropy:

$E(S) = \sum_\limits{i=1}^{n}-p_{i}\log_2p_{i}$

Expanding that summation for the four concept decision attributes for your problem gives

$E(S) = -p_{cinema}\log_2\left(p_{cinema}\right) - p_{tennis}\log_2\left(p_{tennis}\right) - p_{stayin}\log_2\left(p_{stayin}\right) - p_{shopping}\log_2\left(p_{shopping}\right)$

There are three observations where Weather=Sunny. Of those three, one has Decision=Cinema and two have Decision=Tennis. So for Weather=Sunny you have $p_{cinema}=\frac{1}{3}$, $p_{tennis}=\frac{2}{3}$, and other decision probabilities are zero. Plugging those values into the equation above gives

$
\begin{align}
E(S_{Sun}) &amp;= -\frac{1}{3}\log_2\left(\frac{1}{3}\right)-\frac{2}{3}\log_2\left(\frac{2}{3}\right) - \left(0\right)\cdot\log_2\left(0\right) - \left(0\right)\cdot\log_2\left(0\right) \\
           &amp;= 0.38998 + 0.52832 + 0 + 0 \\
           &amp;= 0.918
\end{align}$