Can You do it without calculating K explicitly...

The entries in $K$ are only determined modulo $26$ . Now $K \; = \; \left(\begin{array}{cc} -4 & 3 \\ 9 & 6 \end{array}\right)$ does the job.

Replacing (-4) by $22$ gives the answer of $\boxed{40}$ .

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Without calculating $K$ explicitly...

$566 \; = \; (1\hspace{0.3cm}1)C\left(\begin{array}{c} 13 \\ 5 \end{array}\right) \; = \; (1 \hspace{0.3cm} 1) KP \left(\begin{array}{c} 13 \\ 5 \end{array}\right) \; = \; (1\hspace{0.3cm}1)K\left(\begin{array}{c} 189 \\ 189 \end{array}\right)$

so that $566 = 189(a+b+c+d)$ and so $7(a+b+c+d) \equiv 20 \pmod{26}$ . Multiplying by $15$ we deduce that $a+b+c+d \equiv 14 \pmod{26}$ . This means that $40$ is the only possible answer from the options given.