Find Consistency Of System Specifications 1. p ∧ ¬q = T

2. (q ∧ p) → r = T

3.¬p → ¬r = T

4.(¬q ∧ p) → r = T

From Eq 1, we got p = T and q = F

Now Apply value of P in Eq 3, we get:

$$\begin{array}{cc}
p&¬p&r&¬r&¬p\to r\\ \hline
\color{red}{\text{T}}&\color{red}{\text{F}}&\color{blue}{\text{T}}&\color{red}{\text{F}}&\color{red}{\text{T}}\\
\color{red}{\text{T}}&\color{red}{\text{F}}&\color{blue}{\text{F}}&\color{red}{\text{T}}&\color{red}{\text{T}}\\
\text{F}&\text{T}&\text{T}&\text{F}&\text{F}\\
\text{F}&\text{T}&\text{F}&\text{T}&\text{T}
\end{array}$$
Now there are two possibilities when  ¬p→r is T, and ¬p is F but the r has two separate values.
Is this System consistent or inconsistent?
 A: It's perfectly ok that your set of sentences is consistent if you have 2 different models satisfying your set of sentences (a theory) since consistency has nothing to do with uniqueness of model as referenced here

a consistent theory is one that does not lead to a logical contradiction. The lack of contradiction can be defined in either semantic or syntactic terms. The semantic definition states that a theory is consistent if it has a model, i.e., there exists an interpretation under which all formulas in the theory are true.

In fact any set of tautological sentences such as {$(p=p), (q=q), (r=r)$} can always have different truth values for any propositional sentence $p,q,r$ to stay to be consistent.
But look further about your set of particular sentences, $r$ can only be true from the constraint of your last sentence 4 since the antecedent of your material conditional is true then $r$ has to be true...
A: A set of sentences as consistent iff their conjunction is satisfiable.
(Informally: a consistent system is one whose premises/axioms are coherent in some universe.)
So, in propositional logic, an inconsistent system is one whose conjunction is a contradiction, i.e., whose conjunction is false regardless of the combination of truth values of its atomic propositions.
So, in your exercise, the system is inconsistent iff $$(1 ∧ 2 ∧ 3 ∧ 4) \equiv\bot,$$ i.e., regardless of $(p,q,r)$'s value,  $(1 ∧ 2 ∧ 3 ∧ 4)=$ F,
i.e., every row of $(1 ∧ 2 ∧ 3 ∧ 4)$'s truth table has a False main connective.
Because the main connective $\to$ in your simplified truth table of $(1 ∧ 2 ∧ 3 ∧ 4)$ is True thrice, your system is consistent.
A: For a system to be consistent it must have $\textbf{one}$ outcome that is true, $T$. For a system to be inconsistent it would no true values in the outcome, in other words is all false values, $F$.
In your example we see we have $T$'s in the right most column meaning this system $\textbf{is}$ consistent.
$$\begin{array}{c|c}
p&¬p&r&¬r&¬p\to r\\ \hline
\color{red}{\text{T}}&\color{red}{\text{F}}&\color{blue}{\text{T}}&\color{red}{\text{F}}&\color{red}{\text{T}}\\
\color{red}{\text{T}}&\color{red}{\text{F}}&\color{blue}{\text{F}}&\color{red}{\text{T}}&\color{red}{\text{T}}\\
\text{F}&\text{T}&\text{T}&\text{F}&\text{F}\\
\text{F}&\text{T}&\text{F}&\text{T}&\text{T}
\end{array}$$
For a system to be in consistent, here is an example:
$$\begin{array}{c|c} p & (p\land \lnot p) \\ \hline T & F \\ F &F 
\end{array}$$
The right most column is always $F$ so this system is inconsistent.
Note: If the right most column is always $T$, that is called a tautology.
Also, for these problems I would do a truth table for each one. You won't have that many rows and it would be easy to see if the system is consistent or not.
