How to calculate $p$ value bound for $\chi^2$ test Consider a hypothesis test concerning the variance from a normal population with $H_0: \sigma_2=339.7$ and $H_a: \sigma_2<339.7$. Select bounds on the $p$ value for $n=11$ and test statistic $\chi^2=1.36$.
A) $0.025\leq p\leq0.05$
B) $0.0001\leq p\leq 0.001$
C) $p\leq 0.0001$
D) $0.005\leq p\leq 0.01$
 A: Just to make sure the rational and computations are clear:
To test $H_0: \sigma = 339.7$ against $H_a: \sigma < 339.7,$
one uses the test statistic
$$Q_{obs} = \frac{(n-1)S^2}{\sigma_0^2},$$
where $S$ is the sample standard deviation of your normal
sample of size $n = 11$ and $\sigma_0 = 339.7.$ 
You do not give the numerical value of the sample SD $S,$ but
you report that $Q_{obs} = 1.36.$
You would reject $H_0$ when $Q_{obs}$ is sufficiently small.
In your case that would be when $S$ is 'significantly' smaller than
the null value $333.7.$
Under $H_0$ (that is, assuming $H_0$ to be true), 
$$Q = \frac{(n-1)S^2}{\sigma} \sim \mathsf{Chisq}(n-1).$$
At the fixed significance level $\alpha =0.01 = 1\%,$ you would
reject $H_0$ if $Q_{obs} < q^*,$ where one sees from printed tables
or software that the 'critical value' $q^* = 2.558$ cuts 1% of the
probability from the lower tail of the distribution $\mathsf{Chisq}(10),$
the chi-squared distribution with degrees of freedom $\nu = n -1 = 11 - 1 = 10.$
This distribution has mean $E(Q) = \nu = 10.$ Because $Q_{obs} = 1.35 < 2.558$
you would reject $H_0$ at the 1% level of significance.
From R statistical software,
qchisq(.01, 10)
## 2.558212

The P-value is the probability under $H_0$ of getting a value of $Q < Q_{obs}.$
In general, one cannot find the exact P-value using tables of the chi-squared
distribution because not enough probabilities and critical values are given.
In your specific case, the P-value from R is 0.00069, which is between
the values 0.0001 and 0.001 in one of your answers.
pchisq(1.36, 10)
## 0.0006907683

Presumably your chi-squared table shows critical values close to
0.8889 and 1.4787, which 'bracket' 1.36.
qchisq(c(.0001, .001), 10)
## 0.8889204 1.4787435

The figure below shows the density curve for $\mathsf{Chisq}(10)$ with a
solid vertical black line at $Q_{obs}=1.36.$ The dashed vertical red line
cuts 1% of the area from the lower tail of the curve. The two dotted brown
lines are at the values mentioned above that bracket your observed value 1.36.

