Smallest positive integers k such that there exists a prime P with the property that the six numbers P, P+K,P+2K,P+3K, P+4K, P+5K are all primes Compact way to find the smallest positive integer $K$ such that there exists a prime $P$ with the property that the six numbers $P$, $P+K$,$P+2K$,$P+3K$,$P+4K$,$P+5K$ are all primes. 

Here we can notice that $K$ and $P$ must be greater than $5$ to avoid getting composite numbers. In addition to this, by simply trial and error, I have got $30$ as $K$ and $7$ as $P$.
Is it possible to get a rigorous proof?

We can generalize this problem as follows: 

'Is it possible to find the largest n and the smallest positive integer $K$ such that there exists a prime $P$ with the property that the $n+1$ numbers $P$, $P+K$,$P+2K$,$\cdots$, $P+nK$ are all primes?'
 A: Consider a given positive integer $n \ge 1$, and a corresponding positive integer $K$, for which a prime $P$ exists such that $P$, $P + K$, $P + 2K$, $\ldots$, $P + nK$ are all primes. Note $K$ must be an integral multiple of the primorial $p_j\#$, where the index refers to the prime index (e.g., $p_1 = 2$, $p_2 = 3$, etc.) and $p_j$ is the largest prime $\le n$. Thus, the smallest possible $K$ would be $p_j\#$ itself. This is why the minimum $K$ for your example where $n = 5 = p_3$ is $30$ since $30 = p_3\# = 2(3)(5)$.
To prove $p_j\# \mid K$, first note if $P = p_i$ for some $1 \le i \le j$, then $p_i \mid P + p_i(K)$, so it can't be prime, which means $P \ge p_{j+1}$. Next, assume there's a $p_i$, with $i \le j$, where $p_i \nmid K$. Then $P$, $P + K$, $\ldots$, $P + (p_i - 1)K$ all have different remainders modulo $p_i$ (since if any $2$ were the same, say $P + qK$ and $P + rK$ with $r \gt q$, the difference of $(r - q)K$ must be divisible by $p_i$ but $0 \lt r - q \lt p_i$). Since there are $p_i$ possible remainders from $0$ to $p_i - 1$ among these $p_i$ values, one of them must be $0$. As it must be $\gt p_i$, it can't be prime. This shows the original assumption that $p_i \nmid K$ must be false.
A special case to consider is if $n + 1$ is prime, i.e., it's $p_{j+1}$. If so, then if $p_{j+1} \nmid K$, you must have $P = p_{j+1}$ since, otherwise, one of the $P + iK$ for $1 \le i \le n$ must be a multiple of $p_{j + 1}$ and $\gt p_{j+1}$, so it can't be prime.
Regarding proving there always exists a prime $P$ where, with $K = p_j\#$, you have $P + ik \; \forall \; 1 \le i \le n$ being prime, there's no general method I know of to prove this. Although I suspect such a $P$ always exists, the only thing known for sure now is that a $K$ and $P$ do exist for any positive $n$, as explained in the next paragraph.
As for finding the largest $n$ where there's a $K$, including a smallest such $K$, where $P$, $P + K$, $\ldots$, $P + nK$ are all prime, there is no such maximum $n$. Note the Green–Tao theorem states

... the sequence of prime numbers contains arbitrarily long arithmetic progressions.

