Looking at note C in the last table, it looks like the model is missing main effects and only includes interaction terms which is unusual and usually a bad idea. But we can still recreate the full equation equation:
\begin{align}
\log \mu_{ijk} &= \lambda + \lambda_{ij}^{PS} + \lambda_{ik}^{PA} \\
\mu_{ijk} &= e^{\lambda + \lambda_{ij}^{PS} + \lambda_{ik}^{PA}} \\
\mu_{ijk} &= e^{\lambda}e^{\lambda_{ij}^{PS}}e^{\lambda_{ik}^{PA}}
\end{align}
The parameters are a multiplicative effect on the estimated count (reference). For example, if $\rm{position}=1$ and $\rm{sex}=1$ and $\rm{appraisal}=1$ then the estimated count is:
$$
\mu_{ijk} = e^{4.742}e^{-1.691}e^{0.439} = 32.786
$$
The parameters are zero since they are the reference level for the model. The Constant parameter is used for the count for $\rm{position}=3$, $\rm{sex}=2$, and $\rm{appraise}=3$ which is $e^{4.742} e^0 e^0=114.66$. Notice if you change any of the 3 variables, at least one of the other two parameters will be non-zero.
After going through point 1 and 2, you can see that this interpretation is not quite correct. The model gives us counts and not "likely". So instead, we can say for people in $\rm{position}=3$, more are $\rm{appraise}=2$ than $\rm{appraise}=3$. And for people in $\rm{position}=3$, less are $\rm{appraise}=1$ than $\rm{appraise}=3$. To answer your comment about accuracy, you can use the 95% confidence intervals for the parameters and compute count confidence intervals. There are also other methods (possibly better) for determining statistical significance.
If you're wondering about "likely", one question that people tend to ask would be: Are men in $\rm{position}=3$ more likely to be $\rm{appraise}=1$ than women in $\rm{position}=3$? To answer this, we can look at odds ratios and look at the standard errors there.
The odds of a man in $\rm{position}=3$ being $\rm{appraise}=1$ vs otherwise is $\frac{62}{257 + 198}=0.13$ to 1
The odds of a woman in $\rm{position}=3$ being $\rm{appraise}=1$ vs otherwise is $\frac{45}{149+ 111}=0.17$ to 1
The odds ratio is then $\frac{62 * (149+111)}{45 * (257+198)} = \frac{0.13}{0.17}=0.78$
The standard error is defined in log odds ratio and is approximately
\begin{align}
SE &= \sqrt{\frac{1}{n_{11}} + \frac{1}{n_{10}} + \frac{1}{n_{01}}+ \frac{1}{n_{00}}} \\
&= \sqrt{\frac{1}{62} + \frac{1}{257+198} + \frac{1}{45} + \frac{1}{149+111}} \\
&= 0.21
\end{align}
To combine the two, we need to compute the log odds ratio, adjust for the standard error, then convert out of log scale. This gives a 95% confidence interval of
$$
\text{Lower Bound} = e^{\log{(Odds Ratio)} - 1.96 SE} = e^{\log{0.78} - 1.96 * 0.21} = 0.52
$$
$$
\text{Upper Bound} = e^{\log{(Odds Ratio)} + 1.96 SE} = e^{\log{0.78} + 1.96 * 0.21} = 1.19
$$
Since we're at least 95% significant, we can say that men have 0.78 times (lower) odds than womens' odds.
Finally, take a look at Agresti's Categorical Data Analysis. Chapter 9,10 [3rd ed.] contains lots of information about interpretation and statistical significance for log linear models. It also has a better model selection technique using likelihood ratios rather than looking at a parameter's statistical significance.
